Single Number II

Description

Given an integer array nums where every element appears three times except for one, which appears exactly once. Find the single element and return it.

You must implement a solution with a linear runtime complexity and use only constant extra space.

Example 1:

Input: nums = [2,2,3,2] Output: 3

Example 2:

Input: nums = [0,1,0,1,0,1,99] Output: 99

Constraints:

• 1 <= nums.length <= 3 * 104
• -231 <= nums[i] <= 231 - 1
• Each element in nums appears exactly three times except for one element which appears once.

Code

在 Single Number 中，兩次 XOR 等於 no-op，但是在這題當中，我們想要做的事情是讓三次 XOR 等於 no-op，但這和 XOR 本身的性質不符合，因此我們必須自己設定 counter。

generalize 到如何使 k 次 XOR 等於 no-op，即是 count[j] = (count[j] + 1) % k; 即可。

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        vector<int> count(32, 0);
        for(int i = 0; i < nums.size(); i++) {
            for(int j = 0; j < 32; j++) {
                bool hasBit = (nums[i] & (1 << j));
                if(hasBit) {
                    count[j] = (count[j] + 1) % 3;
                }
            }
        }
 
        int res = 0;
        for(int i = 0; i < 32; i++) {
            if(count[i] > 0) {
                res |= (1 << i);
            }
        }
        return res;
    }
};

Source

• Single Number II - LeetCode