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 getnums[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.
- 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

Given [4, 1, 5, 10]
Return 270

nums = [4, 1, 5, 10] burst 1, get coins 4 * 1 * 5 = 20
nums = [4, 5, 10]    burst 5, get coins 4 * 5 * 10 = 200
nums = [4, 10]       burst 4, get coins 1 * 4 * 10 = 40
nums = [10]          burst 10, get coins 1 * 10 * 1 = 10

Total coins 20 + 200 + 40 + 10 = 270

递归做法：

 1 public class Solution {
 2     /**
 3      * @param nums a list of integer
 4      * @return an integer, maximum coins
 5      */
 6     public int maxCoins(int[] nums) {
 7         // Write your code here
 8         if (nums == null || nums.length == 0)
 9             return 0;
10
11         ArrayList<Integer> list = new ArrayList<Integer>();
12         for (int index = 0; index < nums.length; index++) {
13             list.add(nums[index]);
14         }
15         return helper(list);
16     }
17
18     public int helper(ArrayList<Integer> list) {
19         if (list.size() == 1) {
20             return list.get(0);
21         }
22
23         int max = 0;
24         for (int i = 0; i < list.size(); i++) {
25             int value = 0;
26             if (i == 0) {
27                 value = list.get(0) * list.get(1);
28             } else if (i == list.size() - 1) {
29                 value = list.get(i - 1) * list.get(i);
30             } else {
31                 value = list.get(i - 1) * list.get(i) * list.get(i + 1);
32             }
33             ArrayList<Integer> temp = new ArrayList<Integer>(list);
34             temp.remove(i);
35
36             max = Math.max(max, value + helper(temp));
37         }
38         return max;
39     }
40 }