【2020-02-11】1346. Check If N and Its Double Exist

 

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Easy

Given an array arr of integers, check if there exists two integers N and M such that N is the double of M ( i.e. N = 2 * M).

More formally check if there exists two indices i and j such that :

i != j 0 <= i, j < arr.length arr[i] == 2 * arr[j]

Example 1: Input: arr = [10,2,5,3] Output: true Explanation: N = 10 is the double of M = 5,that is, 10 = 2 * 5. Example 1: Input: arr = [10,2,5,3] Output: true Explanation: N = 10 is the double of M = 5,that is, 10 = 2 * 5.

Example 2: Input: arr = [7,1,14,11] Output: true Explanation: N = 14 is the double of M = 7,that is, 14 = 2 * 7. Example 2: Input: arr = [7,1,14,11] Output: true Explanation: N = 14 is the double of M = 7,that is, 14 = 2 * 7.

Example 3: Input: arr = [3,1,7,11] Output: false Explanation: In this case does not exist N and M, such that N = 2 * M. Example 3: Input: arr = [3,1,7,11] Output: false Explanation: In this case does not exist N and M, such that N = 2 * M.

Constraints:

2 <= arr.length <= 500 -10^3 <= arr[i] <= 10^3

 

 

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有近半月没有作题了，先作道简单题吧。给一个数字数组，判断数组中是否存在一个数是另外一数的2倍大小。

• force暴力拆解法
• map + 两次循环
• set + 一次循环

force暴力拆解法

暴力拆解法就是双次循环，每次对两个值进行判断

• 复杂度分析
  • 时间复杂度：O(N^2)， N是数组长度
  • 空间复杂度：O(N)

/** * @param {number[]} arr * @return {boolean} */ var checkIfExist = function(arr) { for (let i = 0; i < arr.length; i++) { for (let j = 0; j < arr.length; j++) { if (arr[i] == 2 * arr[j] && i !== j) return true } } return false }; 

map

经过使用map，存储arr的值和下标。在第二次循环arr时，经过比较该值的双倍是否在map中存在，若是存在，而且下标不一致，就返回true。

• 复杂度分析
  • 时间复杂度：O(N)， N是数组长度
  • 空间复杂度：O(N)

/** * @param {number[]} arr * @return {boolean} */ var checkIfExist = function(arr) { let map = new Map() for (let i = 0; i < arr.length; i++) { map.set(arr[i], i) } for (let i = 0; i < arr.length; i++) { let double = arr[i] * 2 if (map.has(double) && map.get(double) !== i) return true } return false }; 

set

循环时判断，该值的一半或者2倍是否在set中存在，若是存在，就返回true，若是不在，则使用set来存储该循环中值。

• 复杂度分析
  • 时间复杂度：O(N)， N是数组长度
  • 空间复杂度：O(N)

/** * @param {number[]} arr * @return {boolean} */ var checkIfExist = function(arr) { let set = new Set() for (let i of arr) { if (set.has(2*i) || i % 2 == 0 && set.has(Math.floor(i / 2))) return true set.add(i) } return false };

 

来源

• leetcode