【leetcode】Majority Element

题目如下

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.

第一中方案如下：比较通用的一种解法，也比较笨

public class Solution {
    public int majorityElement(int[] num) {
        HashMap<Integer,Integer> map=new HashMap<>();
        for(int a:num){
            if(map.get(a)!=null){
                int i=map.get(a);
                 map.put(a,i+1);
            }
            else{
                map.put(a,1);
            }
        }
        List<Map.Entry<Integer,Integer>> list =new ArrayList<>();
        list.addAll(map.entrySet());
        Collections.sort(list,new Comparator<Map.Entry<Integer,Integer>>(){
           public int compare(Map.Entry<Integer,Integer> m1,Map.Entry<Integer,Integer> m2){
               return m2.getValue()-m1.getValue();
           } 
        });
        return list.get(0).getKey();
    }
}

public class Solution {
    public int majorityElement(int[] num) {
        HashMap<Integer,Integer> map=new HashMap<>();
        for(int a:num){
            if(map.get(a)!=null){
                int i=map.get(a);
                 map.put(a,i+1);
            }
            else{
                map.put(a,1);
            }
        }
        for(Map.Entry<Integer,Integer> m:map.entrySet()){
            if(m.getValue()>num.length/2){
                return m.getKey();
            }
        }
        return 0;
    }
}

public class Solution {
    public int majorityElement(int[] num) {
        Arrays.sort(num);
        return num[num.length/2];
    }
}

【leetcode】Majority Element

原文：http://blog.csdn.net/exceptional_derek/article/details/43371973