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?

[解题思路]

人生最悲催的事是刚面完试，leetcode就把我面试的题目给刷出来了。。。。

1.O(n) time complexity O(n) space complexity count the ocurrence of every number

public class Solution {
    public int singleNumber(int[] A) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int N = A.length;
        if(N == 0){
            return N;
        }
        
        Map<Integer, Integer> counts = new HashMap<Integer, Integer>();
        for(int i = 0; i < N; i++){
            if(counts.containsKey(A[i])){
                counts.put(A[i], counts.get(A[i]) + 1);
            } else {
                counts.put(A[i], 1);
            }
        }
        
        Iterator<Map.Entry<Integer, Integer>> iterator = counts.entrySet().iterator();
        while(iterator.hasNext()){
            Map.Entry<Integer, Integer> entry = iterator.next();
            if(entry.getValue() != 3){
                return entry.getKey();
            }
        }
        return 0;
    }
}

 

 2.使用O(1)的空间来解决

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

ref:http://www.mitbbs.com/article_t/JobHunting/32547143.html