LeetCode 312. Burst Balloons

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探讨了如何通过动态规划解决气球爆破问题，旨在获得最大收益。文章提出了一种有效的算法，通过构建DP表格逐步求解最优解。

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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

"""
首刷：求最大的乘积

思路1：直接搜索 22case
思路2：将nums直接作为str 进行序列化 30case

题解 dp[i][j] 表示从i到j之间的最大收益

错误的思想： k最先爆 #这样写 比如case[3,1,5,8] -> 取1的时候 [3] , [5,8]被拆开了

正确的思想：k是[i,j]中最后爆的那个


"""
class Solution(object):
    def maxCoins(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        nums = [1] + nums + [1]  #2段补齐

        dp = [[0 for i in xrange(len(nums))] for j in xrange(len(nums))]
        return self.helper(1,len(nums) - 2,dp,nums)   #[1, len(nums) - 2]这个区间内爆破
        
    def helper(self,i,j,dp,nums):
        if i > j:
            return 0

        if dp[i][j] != 0:
            return dp[i][j]
    
        max_one = 0

        for k in xrange(i,j + 1): #[i,j]之间 所有人都可能是最后1个 包括j
            if j + 1 < len(nums) and i - 1 >= 0:
                #print j + 1
                max_one = max(self.helper(i,k - 1,dp,nums) + nums[i - 1] * nums[k] * nums[j + 1] + self.helper(k + 1,j,dp,nums) , max_one)

        dp[i][j] = max_one
        return max_one