leetcode : 312. Burst Balloons : dp矩阵加括号

312. Burst Balloons

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Total Accepted: 7739 Total Submissions: 21379 Difficulty: Hard

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

比较基础的dp，矩阵加括号变形，只不过变形后的坐标注意一下。（几天不写，手生的要死

public class Solution {   public int maxCoins(int[] nums) {int n=nums.length;int[] tnums=new int[n+2];int[][]maxdp=new int[n+2][n+2];System.arraycopy(nums, 0, tnums, 1, n);tnums[0]=1;tnums[n+1]=1;for(int i=n+1;i>=1;i--){for(int j=i;j<=n+1;j++){if(i==j)maxdp[i][j]=0;else{int max=Integer.MIN_VALUE;for(int k=i;k<j;k++){int temp=maxdp[i][k]+maxdp[k+1][j]+tnums[i-1]*tnums[k]*tnums[j];max=max<temp?temp:max;}maxdp[i][j]=max==Integer.MIN_VALUE?0:max;}}}return maxdp[1][n+1];}}