312 - Burst Balloons

绿水兰田

于 2019-07-18 09:57:21 发布

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分类专栏： leetcode 文章标签： leetcode 算法

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Problem

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

Code

public class Solution {
    
    public int maxCoins(int[] nums) {
        int len = nums.length;
        if (len == 0) {
            return 0;
        }
        int[] tmp = new int[len + 2];
        tmp[0] = 1;
        tmp[len + 1] = 1;
        for (int i = 1; i < len + 1; ++i) {
            tmp[i] = nums[i - 1];
        }
        int[][] dp = new int[len + 2][len + 2];
        for (int i = 2; i < len + 2; ++i) {
            for (int j = i; j < len + 2; ++j) {
                for (int k = 1; k < i; ++k) {
                    //System.out.printf("%d %d %d %d\n", k, dp[j][k], dp[k][j - i], nums[j] * nums[k] * nums[j - i]);
                    dp[j][i] = this.max(dp[j][i], dp[j][k] + dp[j - k][i - k] + tmp[j] * tmp[j - k] * tmp[j - i]);
                }
                //System.out.printf("dp[%d][%d]=%d\n", j, i, dp[j][i]);
            }
        }
        return dp[len + 1][len + 1];
    }
    
    private int max(int a, int b) {
        return a > b ? a : b;
    }
}