Problems:
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?
解法一:按位统计0、1出现的次数。使用每个数的每位相加,取模运算,来获得结果。
class Solution { public: int singleNumber(vector<int>& nums) { const int N=sizeof(int)*8; int count[N]; fill_n(&count[0],N,0); for(int i=0;i<nums.size();i++) for(int j=0;j<N;j++) { count[j]+=(nums[i]>>j)&1; count[j]%=3; } //返回结果 int result=0; for(int j=0;j<N;j++) result+=(count[j]<<j); return result; } };
The second solution is need update.