[Swift]LeetCode312. 戳气球 | Burst Balloons

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Given n balloons, indexed from 0 to n-1. Each balloon is painted with a number on it represented by array nums. You are asked to burst all the balloons. If the you burst balloon i you will get nums[left] * nums[i] * nums[right] coins. Here left and right are adjacent indices of i. After the burst, the left and right then becomes adjacent.

Find the maximum coins you can collect by bursting the balloons wisely.

Note:

• You may imagine nums[-1] = nums[n] = 1. They are not real therefore you can not burst them.
• 0 ≤ n ≤ 500, 0 ≤ nums[i] ≤ 100

Example:

Input: [3,1,5,8]
Output: 167 
Explanation: nums = [3,1,5,8] --> [3,5,8] -->   [3,8]   -->  [8]  --> []
             coins =  3*1*5      +  3*5*8    +  1*3*8      + 1*8*1   = 167

---

有 n 个气球，编号为0 到 n-1，每个气球上都标有一个数字，这些数字存在数组 nums 中。

现在要求你戳破所有的气球。每当你戳破一个气球 i 时，你可以获得 nums[left] * nums[i] * nums[right] 个硬币。 这里的 left 和 right 代表和 i 相邻的两个气球的序号。注意当你戳破了气球 i 后，气球 left 和气球 right 就变成了相邻的气球。

求所能获得硬币的最大数量。

说明:

• 你可以假设 nums[-1] = nums[n] = 1，但注意它们不是真实存在的所以并不能被戳破。
• 0 ≤ n ≤ 500, 0 ≤ nums[i] ≤ 100

示例:

输入: [3,1,5,8]
输出: 167 
解释: nums = [3,1,5,8] --> [3,5,8] -->   [3,8]   -->  [8]  --> []
     coins =  3*1*5      +  3*5*8    +  1*3*8      + 1*8*1   = 167

---

188 ms

class Solution {
    func maxCoins(_ nums: [Int]) -> Int {
        if nums.isEmpty {
            return 0
        }
        if nums.count < 2 {
            return nums[0]
        }
        let coinNums = [1] + nums + [1]
        var coins = Array(repeating: Array(repeating: 0, count: coinNums.count), count: coinNums.count)
        let count = coinNums.count
        for i in 2..<count {
            for j in 0..<count-i {
                for k in j+1..<j+i {
                    coins[j][j+i] = max(coins[j][j+i],coins[j][k] + coins[k][j+i] + coinNums[k] * coinNums[j] * coinNums[j+i])
                }
            }
        }
        
        return coins[0][coinNums.count-1]
    }
}