【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;    }}