leetcode 136
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?
数组中只有一个数只出现一次,其余的数都出现两次,找出出现一次的那个数。采用异或运算来处理。
异或运算的性质:任何一个数字异或它自己都等于0。也就是说,如果我们从头到尾依次异或数组中的每一个数字,那么最终的结果刚好是那个只出现依次的数字,因为那些出现两次的数字全部在异或中抵消掉了。
代码如下:
- class Solution {
- public:
- int singleNumber(vector<int>& nums) {
- int n = ;
- for(int i = ; i < nums.size(); i++)
- {
- n ^= nums[i];
- }
- return n;
- }
- };
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