题目要求 Given a non-empty array of integers, every element appears twice except for one. Find that single one. 题目分析及思路 给定一个非空整数数组,除了一个元素只出现一次外,其余元素均出现两次.可以先对数组排序,然后遍历数组,若数组只有一个元素或连续两个元素不相同时则得到返回值,否则则去除这两个元素. python代码 class Solution: def singleNumber(sel…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 异或 字典 日期 [LeetCode] 题目地址:https://leetcode.com/problems/single-number/ Total Accepted: 183838 Total Submissions: 348610 Difficulty: Easy 题目描述 Given a non-empty array of integers…

leetcode 136. Single Number Given an array of integers, every element appears twice except for one. Find that single one. Note: Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? 解题思路: 如果以线性复杂度和…

LeetCode 136. Single Number(只出现一次的数字)…

[LeetCode]Largest Number 解题报告 标签(空格分隔): LeetCode 题目地址:https://leetcode.com/problems/largest-number/#/description 题目描述: Given a list of non negative integers, arrange them such that they form the largest number. For example, given [3, 30, 34, 5, 9], t…

136. Single Number -- Easy 解答 相同的数,XOR 等于 0,所以,将所有的数字 XOR 就可以得到只出现一次的数 class Solution { public: int singleNumber(vector<int>& nums) { int s = 0; for(int i = 0; i < nums.size(); i++) { s = s ^ nums[i]; } return s; } }; 参考 LeetCode Problems' So…

136. Single Number 除了一个数字,其他数字都出现了两遍. 用亦或解决,亦或的特点:1.相同的数结果为0,不同的数结果为1 2.与自己亦或为0,与0亦或为原来的数 class Solution { public: int singleNumber(vector<int>& nums) { if(nums.empty()) ; ; ;i < nums.size();i++) res ^= nums[i]; return res; } }; 137. Single N…

翻译 给定一个整型数组,除了某个元素外其余元素均出现两次. 找出这个仅仅出现一次的元素. 备注: 你的算法应该是一个线性时间复杂度. 你能够不用额外空间来实现它吗? 原文 Given an array of integers, every element appears twice except for one. Find that single one. Note: Your algorithm should have a linear runtime complexity. Could yo…

Given a non-empty array of integers, every element appears twice except for one. Find that single one. Note: Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? Example 1: Input: [2,2,1] Output:…

最原始的方法:先排序,然后从头查找.若nums[i] = nums[i] + 1则为一对相同的数,i = i  + 2,继续判断.若nums[i] != nums[i] + 1,则输出nums[i].需注意不要越界! class Solution { public: int singleNumber(vector<int>& nums) { int t; ) { t = nums[]; } else { sort(nums.begin(), nums.end()); ; i <…