[Leetcode Java] Majority Element, leetcodemajority

[Leetcode Java] Majority Element, leetcodemajority

The questions are as follows:

Given an array of size n, find the majority element. The majority element is the element that appears more⌊ n/2 ⌋Times.

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

Solution 1 is as follows: a more common solution is also stupid

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();    }}

The second solution is as follows: Make full use of the conditions given in the question

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;    }}

The third solution is my search on the Internet: I feel tired after reading it ..

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

WTF ....