今天值接上菜,不閒聊
今日小菜
題號:182 標題:Duplicate Emails 難度: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
題號:169 標題:Majority Element 難度: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
我的程式碼
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;
}
}
邏輯
用HashMap,把出現過的值當key,次數當value,再去尋找value最大得時後,key是多少
這題花費比較久的時間
在HashMap,上次聽大神(it邦可憐小菜鳥不能發文)說有這個東西(HashMap)可以用,剛好選到這一題,就來試試看了
DAY5心得
沒有心得,繼續去拚題目存庫存去