Single Number Given an array of integers, every element appears twice except for one. Find that single one. c++版: class Solution { public: int singleNumber(int arr[] , int length) { int result=arr[0]; for(int i = 1 ; i < length ; ++i) result = result…

[1]LeetCode 136 Single Number 题意:奇数个数,其中除了一个数只出现一次外,其他数都是成对出现,比如1,2,2,3,3...,求出该单个数. 解法:容易想到异或的性质,两个相同的数异或为0,那么把这串数从头到尾异或起来,最后的数就是要求的那个数. 代码如下: class Solution { public: int singleNumber(vector<int>& nums) { ; ;i<nums.size();i++) sum ^= nums[i…

问题: Given an array of integers, every element appears three times except for one. Find that single one. Note: Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?   Single Number I 升级版,一个数组中其它数出现了…

Single Number I : 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? Solution: 解法不少,贴一种: class…

Given an array of integers, every element appears twice except for one. Find that single one. 本题利用XOR的特性, X^0 = X, X^X = 0, 并且XOR满足交换律. class Solution(object): def singleNumber(self, nums): """ :type nums: List[int] :rtype: int ""…

题目描述: 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? Single Number II Given…

Single Number: 1. 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? 代码: class Solution { publi…

Given an array of integers, every element appears three times except for one. Find that single one. Note:Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? 思路: 这个题是Single Number的进阶版,如果其他的数都出现两次而…

I title: 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? 思路:异或 class Solution { public: int…

LintCode 83. Single Number II (Medium) LeetCode 137. Single Number II (Medium) 以下算法的复杂度都是: 时间复杂度: O(n) 空间复杂度: O(1) 解法1. 32个计数器 最简单的思路是用32个计数器, 满3复位0. class Solution { public: int singleNumberII(vector<int> &A) { int cnt[32] = {0}; int res = 0; f…