各位中秋節連假愉快，我今天坐客運，還包車了呢，司機先生只載我一個人，運氣不錯，好啦今天也要繼續努力解題囉

今天的第一題，莫名其妙選到SQL題了呢，題目我真的都是亂選的，看哪個題目看起來比較親切，然後內容感覺上是我寫得出來的，好啦上題目
題號：183 標題：Customers Who Never Order 難度：Easy

題目內容：
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)

用到重新命名欄位跟NOT IN，痾，這題沒甚麼說的，就再來一題吧

題號：283 標題：Move Zeroes 難度：Easy
題目內仍：
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?

我的程式碼

void moveZeroes(int* nums, int numsSize){
    int i =0,count=0,count2,j;
    
     for(i=0;i<numsSize;i++){ //計算多少個0
        if(nums[i]==0){
            count++;
        }
    }
    count2=numsSize-count;//多少個非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){ //當i已經過了非0的idex，全塞0
                nums[i] = 0;
            }
        }
    }
    
    return nums;
}

花比較久的時間
在於怎麼把正確的0的數量補上
其實這題我也是最直觀的解法，先計算非0的各數，然後再把非0的數值填回去，當都填完以後，其他的全部塞0。

DAY3心得
不知不覺第三天來囉，漸入佳境，大家加油阿，晚安