[Swift]LeetCode312. 戳氣球 | 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: 
Output: 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[3,1,5,8]167 
Explanation:

---

有 n 個氣球，編號爲0 到 n-1，每一個氣球上都標有一個數字，這些數字存在數組 nums 中。

如今要求你戳破全部的氣球。每當你戳破一個氣球 i 時，你能夠得到 nums[left] * nums[i] * nums[right] 個硬幣。 這裏的 left 和 right 表明和 i 相鄰的兩個氣球的序號。注意當你戳破了氣球 i 後，氣球 left 和睦球 right 就變成了相鄰的氣球。

求所能得到硬幣的最大數量。

說明:

• 你能夠假設 nums[-1] = nums[n] = 1，但注意它們不是真實存在的因此並不能被戳破。
• 0 ≤ n ≤ 500, 0 ≤ nums[i] ≤ 100

示例:

輸入: 
輸出: 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[3,1,5,8]167 
解釋:

---

188 ms

 1 class Solution {
 2     func maxCoins(_ nums: [Int]) -> Int {
 3         if nums.isEmpty {
 4             return 0
 5         }
 6         if nums.count < 2 {
 7             return nums[0]
 8         }
 9         let coinNums = [1] + nums + [1]
10         var coins = Array(repeating: Array(repeating: 0, count: coinNums.count), count: coinNums.count)
11         let count = coinNums.count
12         for i in 2..<count {
13             for j in 0..<count-i {
14                 for k in j+1..<j+i {
15                     coins[j][j+i] = max(coins[j][j+i],coins[j][k] + coins[k][j+i] + coinNums[k] * coinNums[j] * coinNums[j+i])
16                 }
17             }
18         }
19         
20         return coins[0][coinNums.count-1]
21     }
22 }