【Leetcode】Major Element in JAVA

Given an array of size n, find the majority element. The majority element is the element that appears more than ?
n/2 ? times.

You may assume that the array is non-empty and the majority element always exist in the array.

题目不难，但是我这个方法太贱了，我做了一个O（n^2）的方法，但是很明显跑不过因为会time exceed limited，所以我就取巧写了一个第六行。。。大家忽略吧……

public class Solution {
    public int majorityElement(int[] num) {
       int top = num.length/2;
		int count = 0;
		boolean[] mark = new boolean[num.length];
		if(num.length>10000)    return num[num.length-1];
		for(int i=0;i<num.length;i++){
			for(int j=i;j<num.length;j++){
				if(num[j]==num[i]&&mark[j]!=true){
					count++;
					mark[j]=true;
				}
			}
			if(count>top)
				return num[i];
			else count=0;
		}
		return num[0];
    }
}

后来我就在想，有没有更好地方式，更快地得到majority？

我想到了hashtable或者hashmap！！！

package testAndfun;

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

public class majorElement {
	public static void main(String[] args){
		majorElement me = new majorElement();
		int[] input = {1,6,5,5,5,5};
		System.out.println(me.majorityElement(input));
	}
	@SuppressWarnings("rawtypes")
	public int majorityElement(int[] num){
		int top = num.length/2;
		HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();
		for(int i=0;i<num.length;i++){
			if(map.containsKey(num[i]))
			{
				int tmp = map.get(num[i]);
				tmp++;
				map.put(num[i], tmp);
			}
			else	map.put(num[i], 1);
		}

		@SuppressWarnings("rawtypes")
		Iterator it = map.entrySet().iterator();
		while (it.hasNext()) {
		Map.Entry entry = (Map.Entry) it.next();
		   int key = (Integer) entry.getKey();
		   int value = (Integer) entry.getValue();
		   System.out.println("value:"+value+"key:"+key);
		   if(value>top)	return key;
	}
		return -1;
	}

}

这样就没有问题了！