312. Burst Balloons 难度：hard 类别：分治、动态规划

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本文探讨了一道经典的动态规划问题——气球爆破问题。通过合理的算法设计，实现了一个高效的解决方案。文章详细介绍了问题背景、解题思路及Python程序实现。

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题目：

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:
(1) You may imagine

nums[-1]
 = nums[n] = 1

. They are not real therefore you can not burst them.
(2) 0 ≤ n ≤ 500, 0 ≤ nums[i] ≤ 100

Example:

Given [3, 1, 5, 8]

Return 167

 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

思路：

leetcode把这道题归类为分治，但是我觉得这道题用动态规划去理解更为合适。声明一个二维数组，定义dp[l][h]表示为以序号为l和h为边界的气球串(h>l)，打破l + 1到h - 1之间的所有球所能取得的最大金币数量。考虑在nums[l...h]中，序号为m(l<m<h)的是其中一个气球，把它看作是整个序列中最后一个打破的（不是第一个，这点很关键），就能推导出它的转移方程dp[l][h] = max(dp[l][h], dp[l][m] + dp[m][h] + nums[l]*nums[m]*nums[h])。首先要对nums做些处理，要在它的头尾个插入1，由此dp[0][n+1]则是最终返回的结果。因为最近在学习python，所以这题尝试用python来写程序。

程序：

def maxCoins(nums):
    nums.append(1)
    nums.insert(0,1)
    dp = [[0 for i in range(len(nums))] for i in range(len(nums))]
    n = len(nums)

    for k in range(2,n):
        for l in range(0,n - k):
            h = l + k
            for m in range(l + 1,h):
                dp[l][h] = max(dp[l][h],nums[l] * nums[m] * nums[h] + dp[l][m] + dp[m][h])

    return dp[0][len(nums) - 1]