Dynamic Programming-Targeting/Poke Balloons

Title description:

There are N targets in a shooting range, and each target has a score, which is stored in the score array. The score for hitting the i-th target is score[left] * score[i] * score[right], and the original left and right targets become adjacent targets. The target with a score of 0 cannot be shot. When left does not exist or cannot be shot, the score is score[i] * score[right]. Similarly, right also follows this rule; when left and right do not exist or cannot be shot When, the score is score[i]. Please calculate how many points can be scored for hitting all targets that can be shot? Design algorithm.

Enter a description :

Number of targets n Score of each target   

Output: Score for hitting all targets

Input example:

4

2

3

0

3

Output example:

12

Code:

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int arr[] = new int[n];
		for (int i = 0; i < n; i++) {
			arr[i] = in.nextInt();
		}
		in.close();
		System.out.println(getScore(arr, 0, n - 1));

	}

	public static int getScore(int[] score, int start, int end) {
		if (score == null || start > end) {
			return 0;
		}
		int n = end - start + 1;
		for (int i = start; i < end; i++) {
			if (score[i] == 0) {
				return getScore(score, start, i - 1)
						+ getScore(score, i + 1, end);

			}
		}
		// 将score数组移到新的数组，在前后分别加一个值为1的数
		int[] arr = new int[n + 2];
		for (int i = start + 1; i <= end + 1; i++) {
			arr[i - start] = score[i - 1];
		}
		arr[0] = arr[n + 1] = 1;

		int[][] dp = new int[n + 2][n + 2];
		int m = n + 2;
		for (int k = 2; k < m; k++) {// 左右下标差，k=2表示left和right中间隔1个
			for (int left = 0; left < m - k; ++left) {
				int right = left + k;
				for (int i = left + 1; i < right; ++i) {// 击中第i个靶
					dp[left][right] = Math.max(dp[left][right], arr[left]
							* arr[i] * arr[right] + dp[left][i] + dp[i][right]);
					// 表示最后一个击中的是i号靶时的最大成绩
					// left right中间的分数等于 最后剩i号靶时 i号靶分数是arr[left]* arr[i] *
					// arr[right]
					// left 到 i 号靶之间分数是dp[left][i] i到right之间分数是dp[i][right]
					// 把所有i遍历一遍取最大值
				}
			}
		}
		return dp[0][m - 1];
	}
}

Reference blog: Leetcode 312. Burst Balloons Analysis