[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：https://www.cnblogs.com/strengthen/p/10409173.html 
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

热烈欢迎，请直接点击！！！

进入博主App Store主页，下载使用各个作品！！！

注：博主将坚持每月上线一个新app！！！

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.

---

给出一些不同颜色的盒子，盒子的颜色由数字表示，即不同的数字表示不同的颜色。
你将经过若干轮操作去去掉盒子，直到所有的盒子都去掉为止。每一轮你可以移除具有相同颜色的连续 k 个盒子（k >= 1），这样一轮之后你将得到 k*k 个积分。
当你将所有盒子都去掉之后，求你能获得的最大积分和。

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

class Solution {
    func removeBoxes(_ boxes: [Int]) -> Int {
        var n:Int = boxes.count
        var dp = [[[Int]]](repeating: [[Int]](repeating: [Int](repeating: 0, count: n), count: n), count: n)
        for i in 0..<n
        {
            for k in 0...i
            {
                dp[i][i][k] = (1 + k) * (1 + k)
            }
        }
        for t in 1..<n
        {
            for j in t..<n
            {
                var i:Int = j - t
                for k in 0...i
                {
                    var res:Int = (1 + k) * (1 + k) + dp[i + 1][j][0]
                    for m in (i + 1)...j
                    {
                        if boxes[m] == boxes[i]
                        {
                            res = max(res, dp[i + 1][m - 1][0] + dp[m][j][k + 1])
                        }
                    }
                    dp[i][j][k] = res
                }
            }
        }
        return n == 0 ? 0 : dp[0][n - 1][0]
    }
}