LeetCode 312. Burst Balloons

原题连接在这里：https://leetcode.com/problems/burst-balloons/

题目：

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

    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

题解：

Let dp[l][r]表示扎破(l, r)范围内全部气球得到的最大硬币数，不含边界.

咱们能够想象：最后的剩下一个气球为m的时候，能够得到的分数为 1 * nums[m] * 1.

For all m from l to r, 有 dp[l][r] = max(dp[l][r], dp[l][m] + nums[l] * nums[m] * nums[r]  + dp[m][r]).

dp[l][m] get maximum, burst all ballons between l and m.

dp[m][n] get maximum bust all ballons between m and r.

Now it leaves position l, r and m only. nums[l]*nums[m]*nums[r].

Maintain the maximum.

l与r的跨度k从2开始逐渐增大；

三重循环依次枚举范围跨度k, 左边界l, 中点m, 右边界r = l + k

Time Complexity: O(n^3). Space: O(n^2).

AC Java:

 1 public class Solution {
 2     public int maxCoins(int[] nums) {
 3         if(nums == null || nums.length == 0){
 4             return 0;
 5         }
 6         int len = nums.length+2;
 7         int [] newNums = new int[len];
 8         for(int i = 1; i<len-1; i++){
 9             newNums[i] = nums[i-1];
10         }
11         newNums[0] = 1;
12         newNums[len-1] = 1;
13         
14         int [][] dp = new int[len][len];
15         for(int k = 2; k<len; k++){
16             for(int l = 0; l<len-k; l++){
17                 int r = l+k;
18                 for(int m = l+1; m<r; m++){
19                     dp[l][r] = Math.max(dp[l][r], dp[l][m] + newNums[l]*newNums[m]*newNums[r] +dp[m][r]);
20                 }
21             }
22         }
23         return dp[0][len-1];
24     }
25 }

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