【LeetCode OJ】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?

java code : O(n)时间复杂度， O(n)空间复杂度 ： 哈希

public class Solution {
    public int singleNumber(int[] A) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
         if(A.length == 0)
            return 0;
        HashMap<Integer,Integer> hash = new HashMap<Integer,Integer>();
		for(int a : A)
		{
			if(hash.containsKey(a))
			{
				hash.put(a, hash.get(a) + 1);
			}
			else hash.put(a, 1);
		}
		for(int key : hash.keySet())
		{
			if(hash.get(key) == 1)
				return key;
		}
		return 0;
    }
}

底下转载 http://www.cnblogs.com/feiling/p/3351379.html O(n) 时间复杂度，O(1)空间复杂度

public class Solution {
    public int singleNumber(int[] A) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
         if(A.length == 0)
            return 0;
        int[] cnt = new int[32];
        for(int i = 0; i < A.length; i++)
        {
            for(int j = 0; j < 32; j++)
            {
                if( (A[i]>>j & 1) ==1)
                {
                    cnt[j] = (cnt[j] + 1)%3;
                }
            }
        }
        int res = 0;
        for(int i = 0; i < 32; i++)
        {
            res += (cnt[i] << i);
        }
        cnt = null;
        return res;
    }
}