戳破氣球-LeetCode

思路

dp[i][j] 表示戳破 [i+1…j-1] 號氣球的最大收益。
假設 k 號氣球（i+1 <= k <= j-1）是 [i+1…j-1] 中最後一個被戳破的，則
dp[i][j] = max {for k = range(i+1, j -1) nums[i] * nums[k] * nums[j] + dp[i][k] + dp[k][j]}

代碼：

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