# LeetCode Solution, Medium, 178. Rank Scores

# [178. Rank Scores](https://leetcode.com/problems/rank-scores/)

### 題目敘述

SQL Schema >

    Create table If Not Exists Scores (id int, score DECIMAL(3,2))
    Truncate table Scores
    insert into Scores (id, score) values ('1', '3.5')
    insert into Scores (id, score) values ('2', '3.65')
    insert into Scores (id, score) values ('3', '4.0')
    insert into Scores (id, score) values ('4', '3.85')
    insert into Scores (id, score) values ('5', '4.0')
    insert into Scores (id, score) values ('6', '3.65')

Table: `Scores`

    +-------------+---------+
    | Column Name | Type    |
    +-------------+---------+
    | id          | int     |
    | score       | decimal |
    +-------------+---------+
    id is the primary key for this table.
    Each row of this table contains the score of a game. Score is a floating point value with two decimal places.

Write an SQL query to rank the scores. The ranking should be calculated according to the following rules:

- The scores should be ranked from the highest to the lowest.
- If there is a tie between two scores, both should have the same ranking.
- After a tie, the next ranking number should be the next consecutive integer value. In other words, there should be no holes between ranks.

Return the result table ordered by `score` in descending order.

**Example 1:**

    Input:
    Scores table:
    +----+-------+
    | id | score |
    +----+-------+
    | 1  | 3.50  |
    | 2  | 3.65  |
    | 3  | 4.00  |
    | 4  | 3.85  |
    | 5  | 4.00  |
    | 6  | 3.65  |
    +----+-------+
    Output:
    +-------+------+
    | score | rank |
    +-------+------+
    | 4.00  | 1    |
    | 4.00  | 1    |
    | 3.85  | 2    |
    | 3.65  | 3    |
    | 3.65  | 3    |
    | 3.50  | 4    |
    +-------+------+

#### 題目翻譯

提供一張資料表 `Scores`，需要依照其分數 `scores` 做排序。需要符合以下三個條件： 

1. 分數由高到低排序
2. 同分則同排名
3. 排名的順序是連續的整數，所以不能有缺號

### 解法解析

這題的需求很簡單的想到就是使用 `DENSE_RANK` 的函數，剛好可以符合此需求。額外要注意的是 `RANK` 是另外一個函數，所以在使用 `rank`欄位的時候需要加上引號，才不會造成錯誤。因為這邊 `DENSE_RANK` 會依照 `score` 做排序出 `rank`，所以最後的資料再依照 `rank` 做排序即可。

`DENSE_RANK` 和 `RANK` 的差異可以參考此篇[文章](https://dotblogs.com.tw/jerrytab/2012/12/20/85817)

#### 解法範例

##### MySQL

```sql
# Write your MySQL query statement below
SELECT
    score,
    DENSE_RANK() OVER(
        ORDER BY
            score DESC
    ) AS 'rank'
FROM
    Scores
ORDER BY
    'rank'
```