JavaScript初学者必要了然11个小工夫

作者:ca88

JavaScript 面试中常见算法难题详解

2017/02/20 · JavaScript · 1 评论 · 算法

原来的小说出处: 王下邀月熊_Chevalier   

JavaScript 面试中常见算法问题详解 翻译自 Interview Algorithm Questions in Javascript() {…} 从属于笔者的 Web 前端入门与工程试行。下文提到的累累难题从算法角度并不必需求么困难,但是用 JavaScript 内置的 API 来完结也许需求一番勘察的。

1# Leetcode 367. Valid Perfect Square

Given a positive integer num, write a function which returns True if num is a perfect square else False.

Note: Do not use any built-in library function such as sqrt.

Example 1:
Input: 16
Returns: True

Example 2:
Input: 14
Returns: False

思路:
以256举例
mid = 128 => 128 * 128 > 256 => end = mid = 128;
mid = 64 => 64 * 64 > 256 => end = mid = 64;
mid = 32 => 32 * 32 > 256 => end = mid = 32;
mid = 16 => 16 * 16 = 256 => return true;

以15举例
mid = 8 => 8 * 8 > 15 => end = mid = 8;
mid = 4 => 4 * 4 > 15 => end = mid = 4;
mid = 2 => 2 * 2 < 15 => start = mid = 2; end = 4;
mid = 3 => 3 * 3 < 15 => start = mid = 3; end = 4;
start 1 = 3 1 = 4 = end, while loop end;
start = 3, 3 * 3 != 15 and end = 4, 4 * 4 != 15;
so return false;

public class Solution {
    public boolean isPerfectSquare(int num) {
        if (num < 1) {
            return false;
        }
        long start = 1;
        long end = num;
        while (start   1 < end) {
            long mid = start   (end - start) / 2;
            if (mid * mid == num) {
                return true;
            } else if (mid * mid < num) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if (start * start == num || end * end == num) {
            return true;
        }
        return false;
    }
}

做的有个别题的解题思路

因为好些个的浏览器都和它十三分,你能够在这一个浏览器中采用它。JavaScript被接受的一对一快,因为它是这么的简练,何况使用范围十分广阔。许多技师过去有的时候认为JavaScript是一门“玩具语言”,但是,AJAX步向市镇后显示出了完全相反的一面,它让JavaScript表现出了截然两样的力量和法力。
鉴于这么些发明的产出,技士今后早已得以创设带有桌面应用程序效果的Web应用程序了,那是大有裨益的,因为数量能够更加快地转移。那是某个精制技艺,它们得以帮助初学者更加好地利用JavaScript。JavaScript的采取范围异平时见,何况还或者有那样多的风格,所以它能够有为数十分多的本领。另外,就算它相当多的编制程序方法,不过本人只选择了十一个本领,小编感觉那几个技艺对初学者领悟JavaScript来讲是很好的的源点。
1,在一个数组的最后增多贰个要素
那么些技艺可以让您选拔Length属性在一个数组的末梢增加多个因素,因为Length属性比数组的结尾一个因素的下标多1。这些点子和“push”方法是平等的。比方:

JavaScript Specification

2# Leetcode 270 Closest Binary Search Tree Value

Product of Array Except Self

除本身之外的数组之积
Given an array of n integers where n > 1, nums, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].
Solve it without division and in O(n).
For example, given [1,2,3,4], return [24,12,8,6].

解题思路:拆分法
[A_234 , A_134 , A_124 , A_123 ]=
[1 , A_1 , A_12 , A_123 ]*
[A_234 , A_34 , A_4 , 1 ]

/**
 * Created by Administrator on 2017/5/8.
 */
public class LeetCode {

    public static void main(String[] args) {
        // int [] nums={5, 7, 1, 8,3, 10};  //测试
        int[] nums = {1, 3, 5, 6};
        int k = 5;
        int [] res = productExceptSelf(nums);
        for (int i=0;i<res.length;i  ) {
            System.out.print(res[i] " ");
        }
    }

    public static int[] productExceptSelf(int[] nums) {
        final int [] result = new int [nums.length];
        final int [] right = new int [nums.length];
        final int [] left = new int [nums.length];
        left[0]=1;
        for(int i=1;i<nums.length;i  ){
            left[i]=left[i-1]*nums[i-1];
        }
        right[nums.length-1]=1;
        for(int i=nums.length-2;i>=0;i--){
            right[i]=right[i 1]*nums[i 1];
        }

        for (int i=0;i<nums.length;i  ){
            result[i]=right[i]*left[i];
        }
        return  result;
    }
}

复制代码 代码如下:

演讲下 JavaScript 中的变量提高

所谓升高,看名称就能够想到其意义就是 JavaScript 会将装有的宣示升高到眼下效能域的顶端。那也就代表大家能够在有些变量证明前就选拔该变量,不过即便JavaScript 会将宣示升高到顶端,但是并不会试行真的早先化进程。

3# Leetcode 167 Two Sum II - Input array is sorted

/** 
 *  Method one: Two points 一刷
 *    时间复杂度为O(n), 空间复杂度为O(1)。
 */
public int[] twoSum(int[] numbers, int target) {
    int start = 0;
    int end = numbers.length - 1;
    while (start < end) {
        if (numbers[start]   numbers[end] < target) {
            start   ;
        }
        else if(numbers[start]   numbers[end] > target) {
            end --;
        }
        else {
            break;
        }
    }
    return new int[]{start   1, end   1};
}

/**
 *     Method 2: Binary Search 一刷
 *     时间复杂度为O(logn), 空间复杂度为O(1)。
 */
public class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int[] result = {0,0};
        int index1 = 0;
        int index2 = 0;

        for(int i = 0; i < numbers.length - 1; i   ){
            index1 = i   1;
            if(numbers[i] > target) {
                return result;
            }

            int gap = target - numbers[i];
            int start = i   1;
            int end = numbers.length - 1;

            while(start   1 < end){
                int mid = start   (end - start) / 2;
                if(numbers[mid] == gap) {
                    index2 = mid   1;
                    result[0] = index1;
                    result[1] = index2;
                    return result;
                }
                if (numbers[mid] > gap) {
                    end = mid;
                }
                if (numbers[mid] < gap) {
                    start = mid;
                }
            }
            if (numbers[start] == gap) {
                result[0] = index1;
                result[1] = start   1;
            }
            if (numbers[end] == gap) {
                result[0] = index1;
                result[1] = end   1;
            }
        }
       return result;
    }
}

Increasing Triplet Subsequence

Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array.

Formally the function should:
Return true if there exists i, j, k
such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return false.
Your algorithm should run in O(n) time complexity and O(1) space complexity.

Examples:
Given [1, 2, 3, 4, 5],
return true.

Given [5, 4, 3, 2, 1],
return false.
用整数最大值去比较,x1记录第一个数,x2记录第二大的数,当出现第三大的数,则return true。

public class Solution {
    public boolean increasingTriplet(int[] nums) {
        int x1=Integer.MAX_VALUE;
        int x2=Integer.MAX_VALUE;

        for(int x: nums){
            if(x<=x1) x1=x;
            else if(x<=x2) x2=x;
            else return true;
        }
        return false;
    }
}

var myArray = [];
myArray[myArray.length] = 'New Element';

阐述下 use strict; 的作用

use strict; 看名称就会想到其意义也便是 JavaScript 会在所谓严酷格局下进行,其叁个首要的优势在于可以强制开辟者制止采纳未证明的变量。对于老版本的浏览器依旧推行引擎则会自动忽略该指令。

JavaScript

// Example of strict mode "use strict"; catchThemAll(); function catchThemAll() { x = 3.14; // Error will be thrown return x * x; }

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// Example of strict mode
"use strict";
 
catchThemAll();
function catchThemAll() {
  x = 3.14; // Error will be thrown
  return x * x;
}

4# Leetcode 441. Arranging Coins

You have a total of n coins that you want to form in a staircase shape, where every k-th row must have exactly k coins.

Given n, find the total number of full staircase rows that can be formed.

n is a non-negative integer and fits within the range of a 32-bit signed integer.

Example 1: n = 5
The coins can form the following rows:
¤
¤ ¤
¤ ¤
Because the 3rd row is incomplete, we return 2.

Example 2: n = 8
The coins can form the following rows:
¤
¤ ¤
¤ ¤ ¤
¤ ¤
Because the 4th row is incomplete, we return 3.

思路:

1 2 3 ... k <= n
=>
(k * ( k 1)) / 2 <= n

public class Solution {
    public int arrangeCoins(int n) {
        int start = 0;
        int end = n;
        int mid = 0;
        while (start <= end){
            mid = start   (end - start) / 2 ;
            if ((0.5 * mid * mid   0.5 * mid ) <= n){
                start = mid   1;
            }else{
                end = mid - 1;
            }
        }
        return start - 1;
    }
}

Contains Duplicate II

Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the absolute difference between i and j is at most k.

维护一个HashMap,key为整数,value为下标,将数组中的元素不断添加到这个Hashmap中,遇到重复时,计算下标距离;
用Integer.MAX_VALUE 设置为比较的初始值;
学会用HashMap是非常关键的。

public class LeetCode {
    public static void main(String[] args) {
        int[] nums = {1, 3, 5, 1,6};
        int k=3;
        System.out.print(containsNearbyDuplicate(nums,k));
    }

    public static boolean containsNearbyDuplicate(int[] nums, int k) {
        final Map<Integer,Integer> map = new HashMap<>();
        int min=Integer.MAX_VALUE;

        for(int i=0;i<nums.length;i  ){
            if(map.containsKey(nums[i])){
                final int preIndex=map.get(nums[i]);
                final int gap = i-preIndex;
                min = Math.min(min,gap);
            }
            map.put(nums[i],i);
        }
        return min<=k;
    }
}

2,调解三个数组的长度
Length属性不是只读的,所以你能够安装Length属性的值。何况,你可以行使它增大或缩短数组的长短。比如:

分解下怎么样是 Event Bubbling 以及怎么样防止

Event Bubbling 即指有些事件不唯有会触发当前成分,还有只怕会以嵌套顺序传递到父成分中。直观来说就是对于有些子元素的点击事件同样会被父成分的点击事件管理器捕获。制止Event Bubbling 的措施能够使用event.stopPropagation() 可能 IE 9 以下使用event.cancelBubble

5# Leetcode 35. Search Insert Position

Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

You may assume no duplicates in the array.

Here are few examples.
[1,3,5,6], 5 → 2
[1,3,5,6], 2 → 1
[1,3,5,6], 7 → 4
[1,3,5,6], 0 → 0

public class Solution {
    public int searchInsert(int[] nums, int target) {
        if (nums.length == 0 || nums == null) {
            return 0;
        }
        int start = 0;
        int end = nums.length - 1;
        while (start   1 < end) {
            int mid = start   (end - start) / 2;
            if (nums[mid] == target) {
                return mid;
            }
            else if (nums[mid] < target) {
                start = mid;
            }
            else {
                end = mid;
            }
        }
        if (nums[start] >= target) {
            return start;
        }
        else if (nums[end] >= target) {
            return end;
        }
        else {
            return end   1;
        }
    }
}

Add Two Numbers

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Input: (2 -> 4 -> 3) (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
主要学习怎么创建链表,怎么定义链表

public class LeetCode {
    public static void main(String[] args) {
        int[] inputl1=new int[]{2,4,3};
        int[] inputl2=new int[]{5,6,4};
        ListNode l1=buildListNode(inputl1);
        ListNode l2=buildListNode(inputl2);
        ListNode listNode =addTwoNumbers(l1,l2);
        while(listNode!=null){
            System.out.println("val " listNode.val);
            listNode=listNode.next;
        }
    }
    //定义链表
   public static class ListNode{
        int val;
        ListNode next;
        ListNode(int val){
            this.val=val;
            this.next=null;
        }
    }
    //创建链表
    private static ListNode buildListNode(int[] input){
        ListNode first = null,last = null,newNode;
        int num;
        if(input.length>0){
            for(int i=0;i<input.length;i  ){
                newNode=new ListNode(input[i]);
                newNode.next=null;
                if(first==null){
                    first=newNode;
                    last=newNode;
                }
                else{
                    last.next=newNode;
                    last=newNode;
                }
            }
        }
        return first;
    }

    /**
     * Definition for singly-linked list.
     * public class ListNode {
     *     int val;
     *     ListNode next;
     *     ListNode(int x) { val = x; }
     * }
     */
        public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
           ListNode dummy =new ListNode(-1);
           int carry = 0;
           ListNode prev = dummy;
           for(ListNode pa=l1 ,  pb=l2 ; pa!=null || pb!=null ;
            pa=pa==null?null : pa.next,
            pb=pb==null ? null : pb.next,
            prev=prev.next){
               final int ai=pa==null?0:pa.val;
               final int bi=pb==null?0:pb.val;
               final int value=(ai bi carry);
               carry=(ai bi carry)/10;
               prev.next=new ListNode (value);
            }

            if(carry>0)
                prev.next=new ListNode (carry);
            return dummy.next;
    }
}

复制代码 代码如下:

== 与 === 的界别是什么

=== 也正是所谓的从严比较,关键的分别在于=== 会同一时候相比较类型与值,并不是仅相比值。

JavaScript

// Example of comparators 0 == false; // true 0 === false; // false 2 == '2'; // true 2 === '2'; // false

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// Example of comparators
0 == false; // true
0 === false; // false
 
2 == '2'; // true
2 === '2'; // false

6# Leetcode 374. Guess Number Higher or Lower

We are playing the Guess Game. The game is as follows:

I pick a number from 1 to n. You have to guess which number I picked.

Every time you guess wrong, I'll tell you whether the number is higher or lower.

You call a pre-defined API guess(int num) which returns 3 possible results (-1, 1, or 0):

-1 : My number is lower
1 : My number is higher
0 : Congrats! You got it!

Example:
n = 10, I pick 6.
Return 6.

/* The guess API is defined in the parent class GuessGame.
   @param num, your guess
   @return -1 if my number is lower, 1 if my number is higher, otherwise return 0
      int guess(int num); */

public class Solution extends GuessGame {
    public int guessNumber(int n) {
        int start = 1, end = n;
        while(start   1 < end) {
            int mid = start   (end - start) / 2;
            if(guess(mid) == 0) {
                return mid;
            } else if(guess(mid) == 1) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if(guess(start) == 1) {
            return end;
        }
        return start;
    }
}

Evaluate Reverse Polish Notation

算算逆波兰共和国表明式(又叫后缀表达式)的值

'' 2 '','' 1 '','' '', ''3'', ''* '' -->(2 1)*3-->9

用仓库碰到运算符则把前边多少个拿出去运算

public class Main {
    public static void main(String[] args) {
       String []tokens={"2", "1", " ", "3", "*"};
        System.out.print(evalRPN(tokens));
    }

    public static int evalRPN(String [] tokens){
        Stack<String> s = new Stack<>();
        if(tokens.length==1){
            return Integer.parseInt(tokens[0]);
        }
        for(String token:tokens){
            if(!isOperator(token)){
                s.push(token);
            }else {
                int y=Integer.parseInt(s.pop());
                int x=Integer.parseInt(s.pop());
                switch (token.charAt(0)){
                    case ' ':x =y;break;
                    case '-':x-=y;break;
                    case '*':x*=y;break;
                    case '/':x/=y;break;
                }
                s.push(String.valueOf(x));
            }
        }
        return Integer.parseInt(s.peek());
    }

    private static boolean isOperator(final String op){
        return op.length() == 1 && OPS.indexOf(op)!=-1;
    }

    private static String OPS = new String(" -*/");
}

var myArray = [1,2,3];
myArray.length // 3
myArray.length = 2; //Delete the last element
myArray.length = 20 // add 18 elements to the array; the elements have the undefined value.

解释下 null 与 undefined 的区别

JavaScript 中,null 是贰个方可被分配的值,设置为 null 的变量意味着其无值。而 undefined 则象征着某些变量尽管声称了而是从未开展过别的赋值。

7# Leetcode 69. Sqrt(x)

Implement int sqrt(int x).
Compute and return the square root of x.

public class Solution {
    public int mySqrt(int x) {
        long start = 1;
        long end = x;
        while (start   1 < end) {
            long mid = start   (end - start) / 2;
            if(mid * mid <= x) {
                start = mid;
            }
            else {
                end = mid;
            }
        }
        if(end * end <= x) {
            return (int)end;
        }
        return (int)start;
    }
}

3,使用“!!”把自由数据类型转换到Boolean
以此手艺能够让您采纳“!!”把自由数据类型(举例string, number或integer)调换来Boolean。比如:

解释下 Prototypal Inheritance 与 Classical Inheritance 的区别

在类承继中,类是不可变的,分化的言语中对此多一连的援助也不一致,有些语言中还辅助接口、final、abstract 的概念。而原型承接则越来越灵活,原型自个儿是足以可变的,而且对象大概延续自三个原型。

8# Leetcode 278. First Bad Version

You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad version are also bad.

Suppose you have n versions [1, 2, ..., n] and you want to find out the first bad one, which causes all the following ones to be bad.

You are given an API bool isBadVersion(version) which will return whether version is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.

/* The isBadVersion API is defined in the parent class VersionControl.
      boolean isBadVersion(int version); */

public class Solution extends VersionControl {
    public int firstBadVersion(int n) {
        int start = 1;
        int end = n;
        while (start   1 < end) {
            int mid = start   (end - start) / 2;
            if (isBadVersion(mid)) {
                end = mid;
            }
            else {
                start = mid;
            }
        }
        if(isBadVersion(start)) {
            return start;
        }
        return end;
    }
}

复制代码 代码如下:

数组

9# Leetcode 475. Heaters

Winter is coming! Your first job during the contest is to design a standard heater with fixed warm radius to warm all the houses.

Now, you are given positions of houses and heaters on a horizontal line, find out the minimum radius of heaters so that all houses could be covered by those heaters.

So, your input will be the positions of houses and heaters separately, and your expected output will be the minimum radius standard of heaters.

Note:
Numbers of houses and heaters you are given are non-negative and will not exceed 25000.

Positions of houses and heaters you are given are non-negative and will not exceed 10^9.

As long as a house is in the heaters' warm radius range, it can be warmed.

All the heaters follow your radius standard and the warm radius will the same.

Example 1:
Input: [1,2,3],[2]
Output: 1
Explanation: The only heater was placed in the position 2, and if we use the radius 1 standard, then all the houses can be warmed.

Example 2:
Input: [1,2,3,4],[1,4]
Output: 1
Explanation: The two heater was placed in the position 1 and 4. We need to use radius 1 standard, then all the houses can be warmed.

升序排列加热器的坐标heaters
遍历房子houses,记当前房屋坐标为house:
选取二分查找,分别找到不当先house的最大加热器坐标left,以及不低于house的蝇头加热器坐标right(即左右近日的heater), 则当前房子所需的小不点儿加热器半径radius = min(house - left, right - house)。利用radius更新最后答案。

public class Solution {
    public int findRadius(int[] houses, int[] heaters) {
        //sort
        Arrays.sort(houses);
        Arrays.sort(heaters);

        int radius = 0;
        for( int house: houses) {
            int local = binarySearch(heaters, house);
            radius = Math.max(radius, local);
        }
        return radius;
    }

    private int binarySearch(int[] heaters, int target) {
        int start = 0;
        int end = heaters.length - 1;
        while (start   1 < end) {
            int mid = start   (end - start) / 2;
            if (heaters[mid] == target) {
                return 0;
            } else if (heaters[mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        return Math.min (Math.abs(target - heaters[start]),
                        Math.abs(target - heaters[end]));
    }
}

var myString = '23255';
typeof myString; //String
myString = !!myString;
typeof myString //Boolean

找寻整型数组中乘积最大的四个数

给定三个带有整数的冬天数组,须求寻找乘积最大的八个数。

JavaScript

var unsorted_array = [-10, 7, 29, 30, 5, -10, -70]; computeProduct(unsorted_array); // 21000 function sortIntegers(a, b) { return a - b; } // greatest product is either (min1 * min2 * max1 || max1 * max2 * max3) function computeProduct(unsorted) { var sorted_array = unsorted.sort(sortIntegers), product1 = 1, product2 = 1, array_n_element = sorted_array.length - 1; // Get the product of three largest integers in sorted array for (var x = array_n_element; x > array_n_element - 3; x--) { product1 = product1 * sorted_array[x]; } product2 = sorted_array[0] * sorted_array[1] * sorted_array[array_n_element]; if (product1 > product2) return product1; return product2 };

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var unsorted_array = [-10, 7, 29, 30, 5, -10, -70];
 
computeProduct(unsorted_array); // 21000
 
function sortIntegers(a, b) {
  return a - b;
}
 
// greatest product is either (min1 * min2 * max1 || max1 * max2 * max3)
function computeProduct(unsorted) {
  var sorted_array = unsorted.sort(sortIntegers),
    product1 = 1,
    product2 = 1,
    array_n_element = sorted_array.length - 1;
 
  // Get the product of three largest integers in sorted array
  for (var x = array_n_element; x > array_n_element - 3; x--) {
      product1 = product1 * sorted_array[x];
  }
  product2 = sorted_array[0] * sorted_array[1] * sorted_array[array_n_element];
 
  if (product1 > product2) return product1;
 
  return product2
};

10# Leetcode 349. Intersection of Two Arrays

Given two arrays, write a function to compute their intersection.

Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2].

Note:
Each element in the result must be unique.
The result can be in any order.

public class Solution {
    public int[] intersection(int[] nums1, int[] nums2) {
        if(nums1 == null || nums2 == null) {
            return null;
        }

        HashSet<Integer> set = new HashSet<>();
        Arrays.sort(nums1);

        for (int i = 0; i < nums2.length; i  ) {
            if(set.contains(nums2[i])){
                continue;
            }
            if(binarySearch(num1, nums2[i])) {
                set.add(nums2[i]);
            }
        }

        int[] result = new int[set.size()];
        int index = 0;
        for (Integer num : set) {
            result[index  ] = num;
        }
        return result;
    }

    private boolean binarySearch(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return false;
        }
        int start = 0;
        int end = nums.length - 1;
        while (start   1 < end) {
            int mid = start   (end - start) / 2;
            if (nums[mid] == target) {
                return true;
            }
            else if (nums[mid] < target) {
                start = mid;
            }
            else {
                end = mid;
            }
        }

        if(nums[start] == target || nums[end] == target) {
            return true;
        }
        return false;
    }
}

4,把Number转换成String
以此技巧能够让你在number的终极增添二个空的string来把number调换到string,比方:

搜索接二连三数组中的缺点和失误数

给定某冬日数组,其蕴藉了 n 个一连数字中的 n – 1 个,已知上下面界,要求以O(n)的复杂度寻觅缺点和失误的数字。

JavaScript

// The output of the function should be 8 var array_of_integers = [2, 5, 1, 4, 9, 6, 3, 7]; var upper_bound = 9; var lower_bound = 1; findMissingNumber(array_of_integers, upper_bound, lower_bound); //8 function findMissingNumber(array_of_integers, upper_bound, lower_bound) { // Iterate through array to find the sum of the numbers var sum_of_integers = 0; for (var i = 0; i < array_of_integers.length; i ) { sum_of_integers = array_of_integers[i]; } // 以高斯求和公式计算理论上的数组和 // Formula: [(N * (N 1)) / 2] - [(M * (M - 1)) / 2]; // N is the upper bound and M is the lower bound upper_limit_sum = (upper_bound * (upper_bound 1)) / 2; lower_limit_sum = (lower_bound * (lower_bound - 1)) / 2; theoretical_sum = upper_limit_sum - lower_limit_sum; // return (theoretical_sum - sum_of_integers) }

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// The output of the function should be 8
var array_of_integers = [2, 5, 1, 4, 9, 6, 3, 7];
var upper_bound = 9;
var lower_bound = 1;
 
findMissingNumber(array_of_integers, upper_bound, lower_bound); //8
 
function findMissingNumber(array_of_integers, upper_bound, lower_bound) {
 
  // Iterate through array to find the sum of the numbers
  var sum_of_integers = 0;
  for (var i = 0; i < array_of_integers.length; i ) {
    sum_of_integers = array_of_integers[i];
  }
 
  // 以高斯求和公式计算理论上的数组和
  // Formula: [(N * (N 1)) / 2] - [(M * (M - 1)) / 2];
  // N is the upper bound and M is the lower bound
 
  upper_limit_sum = (upper_bound * (upper_bound 1)) / 2;
  lower_limit_sum = (lower_bound * (lower_bound - 1)) / 2;
 
  theoretical_sum = upper_limit_sum - lower_limit_sum;
 
  //
  return (theoretical_sum - sum_of_integers)
}

11# Leetcode 350. Intersection of Two Arrays II

Given two arrays, write a function to compute their intersection.

Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].

Note:
Each element in the result should appear as many times as it shows in both arrays.
The result can be in any order.
Follow up:
What if the given array is already sorted? How would you optimize your algorithm?
What if nums1's size is small compared to nums2's size? Which algorithm is better?
What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?

public class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        Arrays.sort(nums1);
        Arrays.sort(nums2);
        int index1 = 0;
        int index2 = 0;
        List<Integer> list = new ArrayList<>();
        while(index1 < nums1.length && index2 < nums2.length) {
            if (nums1[index1] == nums2[index2]) {
                list.add(nums1[index1]);
                index1  ;
                index2  ;
            } else if (nums1[index1] < nums2[index2]) {
                index1  ;
            } else if (nums1[index1] > nums2[index2]) {
                index2  ;
            }
        }
        int[] result = new int[list.size()];
        int index = 0;
        for (int element: list) {
            result[index  ] = element;
        }
        return result;
    }
}

复制代码 代码如下:

数组去重

给定某冬辰数组,供给删减数组中的重复数字並且再次来到新的无重复数组。

JavaScript

// ES6 Implementation var array = [1, 2, 3, 5, 1, 5, 9, 1, 2, 8]; Array.from(new Set(array)); // [1, 2, 3, 5, 9, 8] // ES5 Implementation var array = [1, 2, 3, 5, 1, 5, 9, 1, 2, 8]; uniqueArray(array); // [1, 2, 3, 5, 9, 8] function uniqueArray(array) { var hashmap = {}; var unique = []; for(var i = 0; i < array.length; i ) { // If key returns null (unique), it is evaluated as false. if(!hashmap.hasOwnProperty([array[i]])) { hashmap[array[i]] = 1; unique.push(array[i]); } } return unique; }

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// ES6 Implementation
var array = [1, 2, 3, 5, 1, 5, 9, 1, 2, 8];
 
Array.from(new Set(array)); // [1, 2, 3, 5, 9, 8]
 
 
// ES5 Implementation
var array = [1, 2, 3, 5, 1, 5, 9, 1, 2, 8];
 
uniqueArray(array); // [1, 2, 3, 5, 9, 8]
 
function uniqueArray(array) {
  var hashmap = {};
  var unique = [];
  for(var i = 0; i < array.length; i ) {
    // If key returns null (unique), it is evaluated as false.
    if(!hashmap.hasOwnProperty([array[i]])) {
      hashmap[array[i]] = 1;
      unique.push(array[i]);
    }
  }
  return unique;
}

12# Leetcode 153. Find Minimum in Rotated Sorted Array

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

Find the minimum element.

You may assume no duplicate exists in the array.

public class Solution {
    public int findMin(int[] nums) {
        if (nums == null || nums.length == 0) {
            return -1;
        }
        int start = 0;
        int end = nums.length - 1;
        int target = nums[nums.length - 1];

        while (start   1 < end) {
            int mid = start   (end - start) / 2;
            if (nums[mid] <= target) {
                end = mid;
            }
            else {
                start = mid;
            }
        }
        if (nums[start] <= target) {
            return nums[start];
        } else {
            return nums[end];
        }
    }
}

var mynumber = 234;
typeof mynumber; //Number
mynumber = '';
typeof mynumber; //String

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