LeetCode-single-number-ii

题目

题目描述
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?

解题思路

用二进制去模拟三进制，设a是高位，b是低位
三个相同的数(b为1或0)相加,记这个数为c
1.b=1, c+c+c => b=0
2.b=0, c+c+c => b=0

A的卡诺图(状态转移)

AB C	0	1
00	0	0
01	0	1
10	1	0

B的卡诺图(状态转移)

AB C	0	1
00	0	1
01	1	0
10	0	0

总的卡诺图(状态转移)

此时状态	ta	tb	c	a	b
已加上0个c	0	0	0	0	0
已加上0个c	0	0	1	0	1
已加上1个c	0	1	0	0	1
已加上1个c	0	1	1	1	0
已加上2个c	1	0	0	1	0
已加上2个c	1	0	1	0	0

因此可以得出：
a = (ta & ~tb & ~c) | (~ta & tb & c)
b = ~ta & ((~tb & c) | (tb & ~c))

Note：
数组[2,1,3,2,3,3,2]和数组[2,2,2,3,3,3,1]运算结果是一样的，因为位运算是满足交换律的。

Code

public class Solution {
    public int singleNumber(int[] A) {
        int a = 0;
        int b = 0;
        for(int c : A){
            int ta, tb;
            ta = a;
            tb = b;
            a = (ta&~tb&~c) | (~ta&tb&c);
            b = ~ta&((~tb&c)|(tb&~c));
        }
        return a|b;
    }
}