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# 312.Burst-Balloons

## 题目描述

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:
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

## 代码

### Approach #1 DFS

The best score after adding the `i`th balloon is given by:
`nums[left] * nums[i] * nums[right] + dp(left, i) + dp(i, right)`
class Solution {
public int maxCoins(int[] nums) {
// reframe the problem
int n = nums.length + 2;
int[] new_nums = new int[n];
for (int i = 0; i < nums.length; i++) {
new_nums[i + 1] = nums[i];
}
new_nums[0] = new_nums[n - 1] = 1;
int[][] memo = new int[n][n];
return dfs(memo, new_nums, 0, n - 1);
}
private int dfs(int[][] memo, int[] nums, int left, int right) {
if (left + 1 == right) return 0;
if (memo[left][right] > 0) return memo[left][right];
int ans = 0;
for (int i = left + 1; i < right; i++) {
ans = Math.max(ans, nums[left] * nums[i] * nums[right]
+ dfs(memo, nums, left, i)
+ dfs(memo, nums, i, right));
}
memo[left][right] = ans;
return ans;
}
}

### Approach #2 Dynamic Programming Bottom-Up

`dp[left][right] = Math.max(dp[left][right], new_nums[left] * new_nums[i] * new_nums[right] dp[left][i] + dp[i][right]);`
class Solution {
public int maxCoins(int[] nums) {
int n = nums.length + 2;
int[] new_nums = new int[n];
for (int i = 0; i < nums.length; i++) {
new_nums[i + 1] = nums[i];
}
new_nums[0] = new_nums[n - 1] = 1;
int[][] dp = new int[n][n];
for (int left = n - 2; left >= 0; left--) {
for (int right = left + 2; right < n; right++) {
for (int i = left + 1; i < right; i++) {
dp[left][right] = Math.max(dp[left][right],
new_nums[left] * new_nums[i] * new_nums[right] + dp[left][i] + dp[i][right]);
}
}
}
return dp[0][n - 1];
}
}