LeetCode——Single Number
Question
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?
Solution
用异或操作,包含两个的元素,异或操作后都为0,最后就剩下那个single number。
Answer
class Solution {
public:
int singleNumber(vector<int>& nums) {
int res = 0;
for (int i : nums)
res = res ^ i;
return res;
}
};