[LeetCode] Single Number II

Single Number II

 

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?

这个还比较好想，如果某位上1出现的次数为 3*k +1 的话，那么只出现一次的数在该位必定为1咯。

比较坑的是提交几次发现 符号位老是丢掉，不明所以，后来发现是 

const int MASK 这行出现了问题， 应该是 const unsigned int MASK，然后就好了。

class Solution {public:    int singleNumber(int A[], int n) {        // Note: The Solution object is instantiated only once and is reused by each test case.        assert(A&&n>0&&(n-1)%3==0);        unsigned int ans=0;        for(int i=0;i<32;i++)        {            const unsigned int MASK=(1<<i);            int t=0;            for(int j=0;j<n;j++)                t+=(A[j]&MASK)>0?1:0;            ans|=(t%3==0)?0:MASK;        }        return (int)ans;    }};