136. Single Number leetcode做题报告

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?

class Solution {

public:

int singleNumber(vector<int>& nums) {

for(int i=1;i<nums.size();i++){

nums[0]^=nums[i];

}

return nums[0];

}

};

要求O（n）并且不用额外的变量。
非常巧妙的技巧。相同数取异或为0，所0和所有数的异或为本身，所以最后剩下的就是Single Number。

时间： 2024-08-10 11:32:19

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