[leetcode] Single Number III

最新推荐文章于 2020-05-02 20:05:14 发布

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本文介绍了解决LeetCode难题'单数三元素'的方法，通过位运算实现仅使用常数空间复杂度找到数组中出现一次的两个元素。详细解释了算法原理、步骤和代码实现。

From : https://leetcode.com/problems/single-number-iii/

Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.

For example:

Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].

Note:

The order of the result is not important. So in the above example, [5, 3] is also correct.
Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

class Solution {
public:
    vector<int> singleNumber(vector<int>& nums) {
        vector<int> ans(2);
        int u = 0; // sign is the bit differs
        for(int &n : nums) u ^= n;
        for(int i=0; i<32; ++i) {
            if(u&(1<<i)) {
                u = 1<<i;
                break;
            }
        }
        for(int &n : nums) {
            if(u&n) ans[0] ^= n;
            else ans[1] ^= n;
        }
        return ans;
    }
};