leetcode的Single Number II用Hashmap实现出现问题

题目描述

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?

package offer;

import java.util.HashMap;
import java.util.Map;

public class singleNumberII {

public int singleNumber(int[] A) {
    Map<Integer, Integer> map = new HashMap<Integer, Integer>();
    for(int i = 0; i < A.length; i++){
        if(!map.containsKey(A[i])){
            map.put(A[i], 1);
            }else{
                int k = map.get(A[i]);
                if(k == 2){
                    map.remove(A[i]);
                }
                map.put(A[i], ++k);
            }
        }
    return map.keySet().iterator().next();
}

public static void main(String[] args) {
    // TODO Auto-generated method stub
    int[] B = {1,2,2,3,3,2,3,1,1,5};
    singleNumberII ss = new singleNumberII();
    System.out.println(ss.singleNumber(B));
}

}
输出结果应为：5
结果为： 1

Map.put(a[i],k++)应该放在else里面，否则你刚remove掉了，然后就又put进去了

题目要求不允许用多余的存储空间，你直接用map是不行的