4.Single Number && Single Number （II）

Single Number：

1. Given an array of integers, every element appears twice except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

代码：

class Solution {
public:
    int singleNumber(int A[], int n) {
        int result = 0;
        for(int i = 0; i < n; ++i){
            result ^= A[i];
        }
        return result;
    }
};

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?

方法1：（分别计算各位1的个数 mod 3）

class Solution {
public:
    int singleNumber(int A[], int n) {
        int count[32] = {0};
        int result = 0;
        int bit = sizeof(int) * 8;
        for(int i = 0; i < n; ++i) {
            for(int j = 0; j < bit; ++j){
                if((A[i] >> j) & 0x1) count[j] = (count[j] + 1) % 3;
            }
        }
        for(int i = 0; i < bit; ++i){
            result |= (count[i] << i);
        }
        return result;
    }
};

方法2:(三个变量，分别记录出现一次，出现两次，出现三次的值)

class Solution {
public:
    int singleNumber(int A[], int n) {
        int one, two, three;
        one = two = 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;
    }
};