LeetCode 312. Burst Balloons（戳气球）

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本文介绍了一种使用动态规划解决LeetCode上气球爆破问题的方法，旨在通过智慧地爆破气球来获得最大分数。文章详细阐述了算法的具体实现过程，并提供了两种不同的实现代码。

原题网址：https://leetcode.com/problems/burst-balloons/

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

方法：动态规划，定义二维数组coins，coins[a][b]表示把第a个和第b个气球之间（不含a和b）的气球戳烂，最大能得到的分值。

public class Solution {
    public int maxCoins(int[] nums) {
        int[] dpnums = new int[nums.length+2];
        dpnums[0] = 1;
        dpnums[dpnums.length-1] = 1;
        for(int i=0, j=1; i<nums.length; i++, j++) dpnums[j] = nums[i];

        int[][] coins = new int[dpnums.length][dpnums.length];
        for(int i=2; i<dpnums.length; i++) {
            for(int j=0; j+i<dpnums.length; j++) {
                for(int k=j+1; k<j+i; k++) {
                    coins[j][j+i] = Math.max(coins[j][j+i], coins[j][k] + coins[k][j+i] +
                        dpnums[j] * dpnums[k] * dpnums[j+i]);
                }
            }
        }
        return coins[0][dpnums.length-1];
    }
}

另一种实现方式：

public class Solution {
    public int maxCoins(int[] nums) {
        int[] dpnums = new int[nums.length+2];
        dpnums[0] = 1;
        dpnums[dpnums.length-1] = 1;
        System.arraycopy(nums, 0, dpnums, 1, nums.length);
        
        int[][] coins = new int[dpnums.length][dpnums.length];
        for(int i=2; i<dpnums.length; i++) {
            for(int j=i-2; j>=0; j--) {
                for(int k=i-1; k>j; k--) {
                    coins[j][i] = Math.max(coins[j][i], coins[j][k] + dpnums[j] * dpnums[k] * dpnums[i] + coins[k][i]);
                }
            }
        }
        return coins[0][dpnums.length-1];
    }
}