137. Single Number II

137. Single Number II

 

 

Total Accepted: 80477 Total Submissions: 214984 Difficulty: Medium

Given an array of integers, every element appears three times 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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Code:

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int ret = 0;
        int mask = 1;
        while(mask)
        {
            int countOne = 0;   //number of digit 1
            for(int i = 0; i < nums.size(); i ++)
            {
                if(nums[i] & mask)
                    countOne ++;
            }
            if(countOne % 3 == 1)
                ret |= mask;
            mask <<= 1;
        }
        return ret;
    }
};

int singleNumber(int A[], int n)
{
    int one = 0, two = 0;   //每一位出现一次或者三次的其中一种在one中，每一位出现两次和三次的，都在two中。three中保存位出现三次的。
    for (int i = 0; i < n; i++)
    {
        two |= A[i] & one;  //&代表one中是否出现过某一位。
        one ^= A[i];  //^代表当切仅当第奇次出现。 |=  是保存位。
        int three = one & two;  //因为出现了两次的位，one就一定不会出现（异或掉了），所以只有出现三次的位才会在three中出现。
        one &= ~three;  //消除位出现三次的。
        two &= ~three;
    }
    return one;
}