[Swift]LeetCode546. 移除盒子 | Remove Boxes

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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.

---

給出一些不一樣顏色的盒子，盒子的顏色由數字表示，即不一樣的數字表示不一樣的顏色。
你將通過若干輪操做去去掉盒子，直到全部的盒子都去掉爲止。每一輪你能夠移除具備相同顏色的連續 k 個盒子（k >= 1），這樣一輪以後你將獲得 k*k 個積分。
當你將全部盒子都去掉以後，求你能得到的最大積分和。

示例 1：
輸入:

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

輸出:

23

解釋:

[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 }