/* * 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? */ #include <iostream> #include <vector> using namespace std; class Solution { public: int singleNumber(int A[], int n) { int i = 0; int j = 0; int bit[32] = {0}; int singleOne = 0; for(i = 0; i < 32; i++){ for(j = 0; j < n; j++){ bit[i] = bit[i] + ((A[j] >> i) & 1); } singleOne = singleOne | ((bit[i] % 3) << i); } return singleOne; } }; int main(){ Solution solution; int array[10] = {-3, -3, -3, -120, -120, -120, 36, 36, 36, -4}; cout<<solution.singleNumber(array, 10)<<endl; return 1; }
原文地址:http://blog.csdn.net/wyj7260/article/details/39779057