LeetCode: 136 Single Number(easy)

原创于 2019-04-18 10:20:30 发布 · 144 阅读

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本文介绍了一种线性时间复杂度的算法，用于从整数数组中找出仅出现一次的元素，其余元素均出现两次。该算法巧妙地利用了异或运算的特性，无需额外内存开销。

题目：

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?

代码：

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int result = 0;
        for (auto c : nums)
            result ^= c;
        return result;
    }
};