找LeetCode上简单的题目来撑过30天啦(DAY3)

各位中秋节连假愉快，我今天坐客运，还包车了呢，司机先生只载我一个人，运气不错，好啦今天也要继续努力解题罗

今天的第一题，莫名其妙选到SQL题了呢，题目我真的都是乱选的，看哪个题目看起来比较亲切，然後内容感觉上是我写得出来的，好啦上题目
题号：183 标题：Customers Who Never Order 难度：Easy

题目内容：
Suppose that a website contains two tables, the Customers table and the Orders table. Write a SQL query to find all customers who never order anything.
Table: Customers.
+----+-------+
| Id | Name |
+----+-------+
| 1 | Joe |
| 2 | Henry |
| 3 | Sam |
| 4 | Max |
+----+-------+
Table: Orders.
+----+------------+
| Id | CustomerId |
+----+------------+
| 1 | 3 |
| 2 | 1 |
+----+------------+
Using the above tables as example, return the following:
+-----------+
| Customers |
+-----------+
| Henry |
| Max |
+-----------+

我的SQL指令

select Customers.Name Customers  from Customers WHERE Customers.Id NOT IN(SELECT Orders.CustomerId from Orders)

用到重新命名栏位跟NOT IN，痾，这题没甚麽说的，就再来一题吧

题号：283 标题：Move Zeroes 难度：Easy
题目内仍：
Given an integer array nums, move all 0's to the end of it while maintaining the relative order of the non-zero elements.
Note that you must do this in-place without making a copy of the array.

Example 1:
Input: nums = [0,1,0,3,12]
Output: [1,3,12,0,0]
Example 2:
Input: nums = [0]
Output: [0]

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

Follow up: Could you minimize the total number of operations done?

我的程序码

void moveZeroes(int* nums, int numsSize){
    int i =0,count=0,count2,j;
    
     for(i=0;i<numsSize;i++){ //计算多少个0
        if(nums[i]==0){
            count++;
        }
    }
    count2=numsSize-count;//多少个非0
    j=0;
    for(i=0;i<numsSize;i++){ 
        
        if(nums[i] !=  0){
            nums[j] = nums[i]; //把非0塞到前面去
            count2=count2-1;
            j++;
            if(i >= numsSize-count){ //当i已经过了非0的idex，全塞0
                nums[i] = 0;
            }
        }
    }
    
    return nums;
}

花比较久的时间
在於怎麽把正确的0的数量补上
其实这题我也是最直观的解法，先计算非0的各数，然後再把非0的数值填回去，当都填完以後，其他的全部塞0。

DAY3心得
不知不觉第三天来罗，渐入佳境，大家加油阿，晚安