LeetCode: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?

no extra memory solution:

 

class Solution {
public:
    int singleNumber(int A[], int n) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int single_number = 0;
        for (int i = 0; i < 32; ++i) {// assume dealing with int32.
            int bit = 0;
            for (int j = 0; j < n; ++j) {
                bit = (bit + ((A[j] >> i) & 1)) % 3;
            }
            single_number += (bit << i);
        }
        return single_number;
    }
};

 

another solution:

public int singleNumber(int[] A) {
    int[] bv = new int[32];
    for (int i = 0; i < A.length; i++) {
        for (int j = 0; j < 32; j++) {
            bv[j] += (A[i] & (1 << j)) == 0 ? 0 : 1;
        }
    }
    int res = 0;
    for (int i = 0; i < 32; i++) {
        if (bv[i] % 3 != 0) {
            res |= 1 << i;
        }
    }
    return res;
}