题目: 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? 中文(给定一个整数的数组,每个整数都出现了两次,只有一个出现了一次,找到它  建…

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? 题目意思:找出一个数组中的一个不同的元素. 新手想法,先排序,然后循环找出那个前后不一…

给定一个非空整数数组,除了某个元素只出现一次以外,其余每个元素均出现两次.找出那个只出现了一次的元素. 说明: 你的算法应该具有线性时间复杂度. 你可以不使用额外空间来实现吗? 示例 1: 输入: [2,2,1] 输出: 1 示例 2: 输入: [4,1,2,1,2] 输出: 4 利用异或运算符 交换律:a ^ b ^ c <=> a ^ c ^ b 任何数于0异或为任何数 0 ^ n => n 相同的数异或为0: n ^ n => 0 异或:相同为0,不同为1. 异或同一个数两次…

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:…

交换律:a ^ b ^ c <=> a ^ c ^ b 任何数于0异或为任何数 0 ^ n => n 相同的数异或为0: n ^ n => 0 int singleNumber(int* nums, int numsSize) { ]; ;i<numsSize;i++){ result = result ^ nums[i]; } return result; }…

public class Solution { public int SingleNumber(int[] nums) { Dictionary<int, int> dic = new Dictionary<int, int>(); foreach (var n in nums) { if (!dic.ContainsKey(n)) { dic.Add(n, ); } else { dic[n]++; } } ; foreach (var d in dic) { ) { resul…

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:…

题目: 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: 很简单的一题,直接用异或运算解决:连个相同数进行异或结果为0,0与…

题意:数组中每个数字都出现了两次,只有一个出现一次,找出这个数 思路:很明显不能从头到位遍历来找,首先是超时的原因,再次就是这样很没意思·····但是却没想到什么好办法,因为不了解按位异或(XOR).异或就是相同的两个数结果为0,不同的为1.根据交换律我们知道,数组中两两异或的结果就剩最后那一个落单的数了,因为两个相同的数异或结果为0.所以其实这题就很简单了,只是如果没想到或者根本不知道异或就无法好好解题了. 代码: int singleNumber(vector<int>& nums…

1.题目意思:在数组中,只有一个数字只出现了一次 其他的都出现了两次.找出那个只出现一次的数字. //利用位运算 异或 两个相同的数字异或为0 public int singleNumber(int[] nums) { int res = 0 ; for (int i = 0; i < nums.length; i++) { res ^=nums[i]; } return res; } 2.题目意思:数组中,每个数字都出现三次  除了只有一个数字只出现了一次.找出那个只出现一次的数字. // 方…