找到一个数组中只出现一次的元素，使用异或实现。
Runtime: 80 ms, faster than 79.41%
Memory Usage: 40.3 MB, less than 5.04%
/**
* @param {number[]} nums
* @return {number}
*/
var singleNumber = function(nums) {
return nums.reduce((a, b) => a ^ b)
};
/**
* @param {number[]} nums
* @return {number}
*/
var singleNumber = function(nums) {
let res = 0
let len = nums.length
for (let i = 0; i < len; i++) {
res ^= nums[i]
}
return res
};
D51 136. Single Number 题目链接 136. Single Number 题目分析 返回给定数...
136. Single Number https://leetcode.com/problems/single-n...
136. Single Number Given an array of integers, every elem...
136. Single Number Description Given a non-empty array of...
题目链接：136. Single Number 题目描述 Given a non-empty array of i...
Leetcode: 136. Single NumberGiven a non-empty array of in...
Given a non-empty array of integers, every element appear...
Problem Given a non-empty array of integers, every elemen...
问题描述 Given a non-empty array of integers, every element a...
首先哈希表的做法，时间复杂度够，但是空间肯定不行： 思路2：利用异或操作。异或的性质1：交换律a ^ b = b ...
本文标题：136. Single Number
本文链接：https://www.haomeiwen.com/subject/tgnxmktx.html
网友评论