[Swift]LeetCode546. 移除盒子 | Remove Boxes

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：http://www.javashuo.com/article/p-tyvozdbm-me.html 
➤若是连接不是山青咏芝的博客园地址，则多是爬取做者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持做者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

Given several boxes with different colors represented by different positive numbers. 
You may experience several rounds to remove boxes until there is no box left. Each time you can choose some continuous boxes with the same color (composed of k boxes, k >= 1), remove them and get k*k points.
Find the maximum points you can get.

Example 1:
Input:

[1, 3, 2, 2, 2, 3, 4, 3, 1]

Output:

23

Explanation:

[1, 3, 2, 2, 2, 3, 4, 3, 1] 
----> [1, 3, 3, 4, 3, 1] (3*3=9 points) 
----> [1, 3, 3, 3, 1] (1*1=1 points) 
----> [1, 1] (3*3=9 points) 
----> [] (2*2=4 points) 

Note: The number of boxes n would not exceed 100.

---

给出一些不一样颜色的盒子，盒子的颜色由数字表示，即不一样的数字表示不一样的颜色。
你将通过若干轮操做去去掉盒子，直到全部的盒子都去掉为止。每一轮你能够移除具备相同颜色的连续 k 个盒子（k >= 1），这样一轮以后你将获得 k*k 个积分。
当你将全部盒子都去掉以后，求你能得到的最大积分和。

示例 1：
输入:

[1, 3, 2, 2, 2, 3, 4, 3, 1]

输出:

23

解释:

[1, 3, 2, 2, 2, 3, 4, 3, 1] 
----> [1, 3, 3, 4, 3, 1] (3*3=9 分) 
----> [1, 3, 3, 3, 1] (1*1=1 分) 
----> [1, 1] (3*3=9 分) 
----> [] (2*2=4 分)

---

Runtime: 1784 ms

Memory Usage: 21.6 MB

 1 class Solution {
 2     func removeBoxes(_ boxes: [Int]) -> Int {
 3         var n:Int = boxes.count
 4         var dp = [[[Int]]](repeating: [[Int]](repeating: [Int](repeating: 0, count: n), count: n), count: n)
 5         for i in 0..<n
 6         {
 7             for k in 0...i
 8             {
 9                 dp[i][i][k] = (1 + k) * (1 + k)
10             }
11         }
12         for t in 1..<n
13         {
14             for j in t..<n
15             {
16                 var i:Int = j - t
17                 for k in 0...i
18                 {
19                     var res:Int = (1 + k) * (1 + k) + dp[i + 1][j][0]
20                     for m in (i + 1)...j
21                     {
22                         if boxes[m] == boxes[i]
23                         {
24                             res = max(res, dp[i + 1][m - 1][0] + dp[m][j][k + 1])
25                         }
26                     }
27                     dp[i][j][k] = res
28                 }
29             }
30         }
31         return n == 0 ? 0 : dp[0][n - 1][0]
32     }
33 }