【Leetcode】169. Majority Element

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<nums.length;i++){
             if(nums[first]==nums[i]){
                 count++;
             }else{
                 count--;
             }
             if(count==0){
                 first=i;
                 count=1;
             }
         }
         int count2=0;
         for(int i=0;i<nums.length;i++){
             if(nums[first]==nums[i]){
                 count2++;
             }
                 
         }
         return count2>=major?nums[first]:0;
     }