LeetCode 15 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?

分析：考察位运算。

运用三个整型变量，把他们看做二进制数，用这三个数模拟三进制运算。

ones 每一位存放1（余3）出现一次的情况；

twos 每一位存放1（余3）出现两次的情况；

当ones和twos某一位都是1的时候，说明该位1出现了三次，这时候要把ones和twos相应位置清0，清0的具体做法就是将ones和twos按位求交再取反，之后分别与ones和twos按位求交即可。

这样，最后ones就是结果。

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