LeetCode 312. Burst Balloons

原題連接在這裏：https://leetcode.com/problems/burst-balloons/

題目：

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

題解：

Let dp[l][r]表示扎破(l, r)範圍內全部氣球得到的最大硬幣數，不含邊界.

咱們能夠想象：最後的剩下一個氣球爲m的時候，能夠得到的分數爲 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與r的跨度k從2開始逐漸增大；

三重循環依次枚舉範圍跨度k, 左邊界l, 中點m, 右邊界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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