标签：style blog io ar color os sp for strong
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?
要求O(n)算法，还不能有辅助空间。开始用了各种O(n^2)的方法，都超时，后来突然顿悟了。异或可以消掉两个相同数字。
直接把所有数字异或在一起，就是单独的那个数字了。
#include <iostream> #include <vector> #include <algorithm> using namespace std; class Solution { public://异或可以把两个相同的数字消除 O（n) AC int singleNumber3(int A[], int n) { int ans = 0; for(int i = 0; i < n; i++) { ans ^= A[i]; } return ans; } }; int main() { Solution s; int a[9] = {0,0,1,1,2,3,2,3,4}; int n = s.singleNumber3(a, 9); return 0; }
【leetcode】Single Number (Medium) ☆
标签：style blog io ar color os sp for strong
原文地址：http://www.cnblogs.com/dplearning/p/4110566.html
