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## Problem

You are given n balloons, indexed from 0 to n - 1. Each balloon is painted with a number on it represented by an array nums. You are asked to burst all the balloons.

If you burst the ith balloon, you will get nums[i - 1] * nums[i] * nums[i + 1] coins. If i - 1 or i + 1 goes out of bounds of the array, then treat it as if there is a balloon with a 1 painted on it.

Return the maximum coins you can collect by bursting the balloons wisely.

Example 1:

Input: nums = [3,1,5,8]
Output: 167
Explanation:
nums = [3,1,5,8] --> [3,5,8] --> [3,8] -->  --> []
coins =  3*1*5    +   3*5*8   +  1*3*8  + 1*8*1 = 167


Example 2:

Input: nums = [1,5]
Output: 10


## Code

class Solution {
public int maxCoins(int[] nums) {
if(nums == null || nums.length == 0) return 0;
int n = nums.length;
int[][] dp = new int[n][n];

for(int len = 1; len <= n; len++){
for(int i = 0; i <= n - len; i++){
int j = i + len - 1;

for(int k = i; k <= j; k++){
int left = 1;
int right = 1;

if(i != 0){
left = nums[i - 1];
}

if(j != n - 1){
right = nums[j + 1];
}

int prev = 0;
int next = 0;

if(k != i){
prev = dp[i][k - 1];
}

if(k != j){
next = dp[k + 1][j];
}

dp[i][j] = Math.max(dp[i][j], left * nums[k] * right + prev + next);
}
}
}
return dp[n - 1];
}
}