312. Burst Balloons

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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

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 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

解法1：

C++

1

Java

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public 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]; // record the largest coin for a burst between [i, j]
        for (int len = 1; len <= n; len++) {
            for (int start = 0; start <= n - len; start++) {
                int end = start + len - 1;
                // dp[start][end]
                int max = Integer.MIN_VALUE;
                for (int i = start; i <= end; i++) {
                    int coins = 0;
                    coins = nums[i] * (start == 0 ? 1 : nums[start - 1]) * (end == n - 1 ? 1 : nums[end + 1]);
                    // burst nums[i]
                    if (i != start) {
                        coins += dp[start][i - 1];
                    }
                    if (i != end) {
                        coins += dp[i + 1][end];
                    }
                    max = Math.max(max, coins);
                }
                dp[start][end] = max;
            }
        }
        return dp[0][n - 1];
    }
}