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.

Tips：给定一个大小为n的数组，找到数组中出现次数大于n/2的数字。

思路：解法一：将数组排序后，位于n/2位置的数字就是大于半数的数字。

package easy;import java.util.Arrays;import java.util.Collections;public class L169MajorityElement {         public int majorityElement(int[] nums) {         int len=nums.length;         int low=0;         int high=len-1;         int mid=low+(high-low)/2;         Arrays.sort(nums);         int bignum=nums[mid];         return bignum;                    }     public static void main(String[] args) {        L169MajorityElement l169 = new L169MajorityElement();        int[] nums={1,2,2,2,2,3};        int bignum=l169.majorityElement(nums);        System.out.println(bignum);    }}

 解法二：按顺序遍历数组，第后一个数字等于前一个数字，count++.否则count--;

当count=0时，就需要更换数字。

public int majorityElement2(int[] nums) {         int major=nums.length/2;         int first=0;int count=1;         for(int i=1;i
     
      =major?nums[first]:0;     }