LeetCode - Single Number II

Single Number II

2014.1.13 20:25

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?

Solution:

　　Still, exclusive OR is a wonderful operator.

　　This time every number appears 3 times except an extra number. That is to say, every '1' bit appears 3k times if that extra number has '0' on this bit, or 3k+1 times if it has '1' on this bit. If we can use this 3k property, we can separate that extra number from those triads.

　　Here we define four variables:

　　　　a1: the bits that have appeared 3k+1 times.

　　　　a2: the bits that have appeared 3k+2 times.

　　　　b: the bits that have (or have not) appeared 3k times.

　　The algorithm works in the following process:

　　　　1. initialize all to 0.

　　　　2. for each A[i], update a2 from a1 and A[i].

　　　　3. for each A[i], update a1 from A[i].

　　　　4. for each A[i], update b from a1 and a2.

　　　　5. if a bit appears 3k times, it can't appear 3k+1 or 3k+2 time, thus subtract b from a1 and a2.

　　Time complexity is O(n), space complexity is O(1).

Accepted code:

// 1CE, 1AC, just try to avoid spelling error, OK?
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.
        if(n <= 0){
            return 0;
        }
        
        int i;
        int a1, a2;
        int b;
        
        a1 = a2 = b = 0;
        for(i = 0; i < n; ++i){
            // notice: not a2 = (a1 & A[i]);
            // 1CE here, not a[], but A[]
            a2 |= (a1 & A[i]);
            a1 = (a1 ^ A[i]);
            b = ~(a1 & a2);
            a1 &= b;
            a2 &= b;
        }
        // all bits of '1' appear for 3k + 1 or 3k times, thus the remaining one is the result.
        return a1;
    }
};