【LEETCODE】312-Burst Balloons

Given n balloons, indexed from0 ton-1. Each balloon is painted with a number on it represented by arraynums. You are asked to burst all the balloons. If the you burst ballooni you will getnums[left] * nums[i] * nums[right] coins. Hereleft andright are adjacent indices ofi. After the burst, theleft andright 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

参考：

http://www.hrwhisper.me/leetcode-burst-balloons/

我们可以想象：最后的剩下一个气球为i的时候，可以获得的分数为：nums[-1]*nums[i]*nums[n].

那么介于i,j之间的x，有： 

dp[i][j]=max(dp[i][j],DP(i,x-1)+nums[i-1]*nums[x]*nums[j+1]+DP(x+1,j));

小白＋自学＝都快怀疑自己的智商了 不开森

class Solution(object):    def maxCoins(self, nums):        """        :type nums: List[int]        :rtype: int        """                n=len(nums)        nums=[1]+nums+[1]        dp=[[0 for j in range(n+2)] for i in range(n+2)]              #                def DP(i,j):            if dp[i][j]>0:                return dp[i][j]            for x in range(i,j+1):       #在 i，j 这个当前状态中，x为最后剩下的那个数                 dp[i][j]=max(dp[i][j],DP(i,x-1)+nums[i-1]*nums[x]*nums[j+1]+DP(x+1,j))   #矩阵存着最大值of 把x爆破后的分＋x被剩下前的状态                        return dp[i][j]                        return DP(1,n)            #最终要求的就是状态就是 1，n