[leetcode] Single Number
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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:使用位运算来解决,一个整数和它本身异或之后得到值是0,0与其他整数异或得到的是这个整数本身
输入示例:
5
1 2 1 2 3
代码:
#include<iostream>using namespace std;int singleNumber(int A[], int n){ int res; for(int i=0;i<n;i++){ res=res^A[i]; } return res;}int main(){ int n; cin>>n; int A[n]; for(int i=0;i<n;i++){ cin>>A[i]; } int res; res=singleNumber(A,n); cout<<res<<endl; return 0;}
0 0
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