leetcode submission/20161006(Single number)

最新推荐文章于 2025-09-05 23:16:08 发布

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本文提供了一种使用XOR操作解决寻找数组中唯一出现一次元素的方法。该算法具备线性时间复杂度且不使用额外内存，符合高效查找的要求。

MD假期要没了啊FUCKING SHIT。。。。

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?

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my answer:

class Solution {
public:
int singleNumber(vector<int>& nums) {
if (nums.size() == 1)
return nums.front();
else {
bool first_element = true;
int tmp = nums.front();
for (auto & x : nums) {
if (first_element == true)
first_element = false;
else
tmp ^= x;
}
return tmp;
}
}
};

this algorithm uses XOR as key method to filter the elements in the vector.

For anyone who didn't understood why this works here is an explanation. This XOR operation works because it's like XORing all the numbers by itself. So if the array is {2,1,4,5,2,4,1} then it will be like we are performing this operation

((2^2)^(1^1)^(4^4)^(5)) => (0^0^0^5) => 5.

Hence picking the odd one out ( 5 in this case).