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312. burst balloons

admin 11月 13, 2020 0

Hard

https://leetcode.com/problems/burst-balloons/

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:

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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

方法：

边界条件好难写…

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class (object):
    def maxCoins(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if len(nums)==0:
            return 0
        if len(nums)==1:
            return nums[0]
        
        arr = [0]*(len(nums)+2)
        for i in range(len(nums)):
            arr[i+1] = nums[i]
        arr[0], arr[-1] = 1, 1
        
        dp = [[0]*(len(nums)+2) for _ in range(len(nums)+2)]
        for len_ in range(1, len(nums)+1):
            for left in range(1, len(nums)+2-len_):
                right = left + len_ - 1
                for k in range(left, right+1):
                    dp[left][right] = max(dp[left][right], dp[left][k-1]+dp[k+1][right]+arr[k]*arr[left-1]*arr[right+1])
        return dp[1][len(nums)]
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