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;        // cache the results of dp        int[][] memo = new int[n][n];        // find the maximum number of coins obtained from adding all balloons from (0, len(nums) - 1)        int ans = 0;        // manually burst the last balloon because it has special rules        for(int i = 1; i < n; ++i){            ans = Math.max(ans, new_nums[i] + dp(memo, new_nums, i, n - 1) + dp(memo, new_nums, 0, i));        }        return ans;    }    public int dp(int[][] memo, int[] nums, int left, int right) {        // no more balloons can be added        if (left + 1 == right) return 0;        // we've already seen this, return from cache        if (memo[left][right] > 0) return memo[left][right];        // add each balloon on the interval and return the maximum score        int ans = 0;        for (int i = left + 1; i < right; ++i)            ans = Math.max(ans, nums[left] * nums[right]            + dp(memo, nums, left, i) + dp(memo, nums, i, right));        // add to the cache        memo[left][right] = ans;        return ans;    }}輸入：[1, 2, 3, 4][5, 7, 8]輸出：2056