Single Number
2014-01-28 00:23
381 查看
Single Number
Total Accepted: 10122 TotalSubmissions: 22355My Submissions
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?
Discuss
XOR 位操作，因为出现两次，所有XOR两次正好抵消，最后剩下只出现一次的那个数。
public class Solution {
public int singleNumber(int[] A) {
int bit = 0;
for(int i = 0; i < A.length; i++){
bit ^= A[i];
}
return bit;
}
}
内容来自用户分享和网络整理，不保证内容的准确性，如有侵权内容，可联系管理员处理 
相关文章推荐
- [LeetCode] Single Number
- Single Number
- [LeetCode] Single Number
- Single Number
- [LeetCode] Single Number
- 好题 BUPT OJ84 SINGLE NUMBER
- LeetCode题解：Single Number
- 【LEETCODE OJ】Single Number
- leetcode题目：Single Number和Single Number II
- Single Number
- leetcode || Single Number
- LeetCode | Single Number
- [LeetCode] Single Number
- leetCode OJ（1） Single Number
- LeetCode - Single Number
- LeetCode——Single Number（找出数组中只出现一次的数）
- LeetCode题解：Single Number
- [LeetCode] Single Number 单独的数字
- Single Number
- LeetCode：Single Number