LeetCode[136]Single Number

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/* LeetCode[136]: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?
 */

问题：给你一组数一个数字出现一次，其他的数字出现两次，找出那个出现一次的数字
分析：相同数字异或为0，所以将所有数字都异或后剩下的就是出现一次的数 

 1 int singleNumber(int *nums, int numsSize)
 2 {
 3     int single = 0;
 4     int i = 0;
 5     for(i=0; i<numsSize; i++)
 6     {
 7         single^=nums[i];
 8     }
 9     return single;
10 }

LeetCode[136]Single Number

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原文地址：http://www.cnblogs.com/shiddong/p/4510083.html