Leetcode: 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?

O(N)时间的话，不考虑空间，可以用hash。试了一下STL中的unordered_map，可以通过。看来without using extra memory不是强制性的。

#include <unordered_map>
using namespace std;

class Solution {
public:
    int singleNumber(int A[], int n) {
       unordered_map<int, int> numbers;
       unordered_map<int, int>::iterator iter;
       for (int i = 0; i < n; ++i) {
           if ((iter = numbers.find(A[i])) != numbers.end()) {
               ++iter->second;
           }
           else {
               numbers[A[i]] = 1;
           }
       }
       
       for (iter = numbers.begin(); iter != numbers.end(); ++iter) {
           if (iter->second != 3) {
               return iter->first;
           }
       }
       
       return 0;
    }
};

下面的是大牛们的思路，统计各位上1出现的次数，出现3次必定是3的整数倍。

class Solution {
public:
    int singleNumber(int A[], int n) {
       int count[32] = {0};
       int result = 0;
       for (int i = 0; i < 32; ++i) {
           for (int j = 0; j < n; ++j) {
               count[i] += (A[j] >> i) & 1;
           }
           result |= (count[i] % 3) << i;
       }
       
       return result;
    }
};

更牛的：

class Solution {
public:
    int singleNumber(int A[], int n) {
       int one = 0;
       int two = 0;
       int mask = 0;
       for (int i = 0; i < n; ++i) {
           two |= one & A[i];
           one ^= A[i];
           mask = ~(one & two);
           one &= mask;
           two &= mask;
       }
       
       return one;
    }
};

不过三种方式，速度相差不大。

===================第二次==============

第三种方法，还得写半天啊，虽然知道思路。当做三进制数，如果某位上1出现次数达到三个则丢弃，最终剩下的就是只出现一次的数。

class Solution {
public:
    int singleNumber(int A[], int n) {
       int one = 0;
       int two = 0;
       int three = 0;
       for (int i = 0; i < n; ++i) {
           two |= one & A[i];
           one ^= A[i];
           three = one & two;
           one &= ~three;
           two &= ~three;
       }
       
       return one;
    }
};