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?

共3n+1个数, 只有一个是single, 其余的数都是三个相同的

如 1 1 1 2 2 2 4 5 5 5 

需要找出4

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

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

此解法是参考大神的, 确实高明, 二进制模拟三进制

t1 = A^B 余2位, 称余位, 如果是1, mod2=1

此题甚是巧妙

ones 存储 mod3 = 1

twos  存储 mod3 =2

threes 是 mod3 = 0 

threes求法 ones & twos 即两者都为1才是1, 即2+1

然后取反, 目的是更新ones和twos中的数位

class Solution {
public:
    int singleNumber(int A[], int n) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
              map <int, int> mp;
              map <int, int>::iterator it;
              for(int i = 0;i < n;i++){
                   it = mp.find(A[i]);
                   if(it == mp.end()){
                        mp[A[i]] = 1;
                   }
                   else
                   mp[A[i]]++;
              }
              for(it = mp.begin();it != mp.end();it++){
                  if((*it).second != 3)
                  return (*it).first;
              }
     }
};