LeetCode 137. Single Number II

137. Single Number II

Difficulty: Medium

Given a non-empty array of integers, every element appears three times except for one, which appears exactly once. Find that single one.

Note:

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

Example 1:

1 2	Input: [2,2,3,2] Output: 3

Example 2:

1 2	Input: [0,1,0,1,0,1,99] Output: 99

Solution

Language: Java

数bit

class Solution {
    public int singleNumber(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        int[] bits = new int[32];
        for (int n : nums) {
            int bit = 1;
            for (int i = 0; i < 32; i++) {
                if ((bit & n) != 0) {
                    bits[i]++;
                }
                bit <<= 1;
            }
        }
        int res = 0;
        int bit = 1;
        for (int i = 0; i < bits.length; i++) {
            if (bits[i] % 3 != 0) {
                res += bit;
            }
            bit <<= 1;
        }
        return res;
    }
}

位运算法

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

The code seems tricky and hard to understand at first glance.
However, if you consider the problem in Boolean algebra form, everything becomes clear.

What we need to do is to store the number of ‘1’s of every bit. Since each of the 32 bits follow the same rules, we just need to consider 1 bit. We know a number appears 3 times at most, so we need 2 bits to store that. Now we have 4 state, 00, 01, 10 and 11, but we only need 3 of them.

In this solution, 00, 01 and 10 are chosen. Let ‘ones’ represents the first bit, ‘twos’ represents the second bit. Then we need to set rules for ‘ones’ and ‘twos’ so that they act as we hopes. The complete loop is 00->10->01->00(0->1->2->3/0).

For ‘ones’, we can get ‘ones = ones ^ A[i]; if (twos == 1) then ones = 0’, that can be tansformed to ‘ones = (ones ^ A[i]) & ~twos’.
Similarly, for ‘twos’, we can get ‘twos = twos ^ A[i]; if (ones* == 1) then twos = 0’ and ‘twos = (twos ^ A[i]) & ~ones’. Notice that ‘ones*’ is the value of ‘ones’ after calculation, that is why twos is
calculated later.

Here is another example. If a number appears 5 times at most, we can write a program using the same method. Now we need 3 bits and the loop is 000->100->010->110->001. The code looks like this:

int singleNumber(int A[], int n) {
	int na = 0, nb = 0, nc = 0;
	for(int i = 0; i < n; i++){
		nb = nb ^ (A[i] & na);
		na = (na ^ A[i]) & ~nc;
		nc = nc ^ (A[i] & ~na & ~nb);
	}
	return na & ~nb & ~nc;
}

Or even like this:

int singleNumber(int A[], int n) {
	int twos = 0xffffffff, threes = 0xffffffff, ones = 0;
	for(int i = 0; i < n; i++){
		threes = threes ^ (A[i] & twos);
		twos = (twos ^ A[i]) & ~ones;
		ones = ones ^ (A[i] & ~twos & ~threes);
	}
	return ones;
}

I hope all these above can help you have a better understand of this problem.