找LeetCode上简单的题目来撑过30天啦(DAY5)

今天值接上菜，不闲聊

今日小菜
题号：182 标题：Duplicate Emails 难度：Easy

SQL Schema
Write a SQL query to find all duplicate emails in a table named Person.

For example, your query should return the following for the above table:

Note: All emails are in lowercase.

我的SQL指令

select Email from (select Email,count(*) c from Person group by Email ) as table1 where table1.c>1

今日主菜
题号：169 标题：Majority Element 难度：Easy

Given an array nums of size n, return the majority element.
The majority element is the element that appears more than ⌊n / 2⌋ times. You may assume that the majority element always exists in the array.

Example 1:
Input: nums = [3,2,3]
Output: 3
Example 2:
Input: nums = [2,2,1,1,1,2,2]
Output: 2

Constraints:
• n == nums.length
• 1 <= n <= 5 * 104
• -231 <= nums[i] <= 231 - 1

我的程序码

import java.util.HashMap;

class Solution {
    public int majorityElement(int[] nums) {
        HashMap<Integer,Integer> temp = new HashMap<Integer, Integer>();
        int l = nums.length,i,j,max=0,vmax=0;
        //System.out.print(l);
        for(i=0;i<l;i++){
     

            if(temp.containsKey(nums[i])){
                temp.replace(nums[i], temp.get(nums[i])+1) ;
   
            }else{
                temp.put(nums[i],1);
            }
        }
        Set s =temp.keySet();
        //System.out.println(temp);

         for(Object key : s){
            if(temp.get(key)>vmax){
                 max = (int)key;
                 vmax = temp.get(key);
                //System.out.println(key + ","+temp.get(key));
            }
        }
        
        return max;
    }
}

逻辑
用HashMap，把出现过的值当key，次数当value，再去寻找value最大得时後，key是多少

这题花费比较久的时间
在HashMap，上次听大神(it邦可怜小菜鸟不能发文)说有这个东西(HashMap)可以用，刚好选到这一题，就来试试看了

DAY5心得
没有心得，继续去拚题目存库存去