原题链接在这里：https://leetcode.com/problems/remove-boxes/
题目：
Given several boxes with different colors represented by different positive numbers.
You may experience several rounds to remove boxes until there is no box left. Each time you can choose some continuous boxes with the same color (composed of k boxes, k >= 1), remove them and get k*k points.
Find the maximum points you can get.
Example 1:
Input:
[1, 3, 2, 2, 2, 3, 4, 3, 1]
Output:
23
Explanation:
[1, 3, 2, 2, 2, 3, 4, 3, 1] ----> [1, 3, 3, 4, 3, 1] (3*3=9 points) ----> [1, 3, 3, 3, 1] (1*1=1 points) ----> [1, 1] (3*3=9 points) ----> [] (2*2=4 points)
题解：
Let T(l, r, k) denotes maximum points obtained within boxes[l~r], and there are k boxes on the right side of boxes[r] having same color as boxes[r].
There could be 2 cases:
case 1: remove boxes[r] with another k on the right of it and get (k+1)*(k+1) points. The rest is T(l, r-1, 0).
case 2: Not remove boxes[r] for now. And for each i between l~r-1, if boxes[i] has same color as boxes[r]. divide it into 2 parts. T(l, i, k+1), r is accumlated to k+1. And T(i+1, r-1, 0).
Some optimization is to to move r to left and increase k if boxes[r] and boxes[r-1] have same color. Save time to calculate remove boxes[r] because it must obintain more points remove boxes[r] and boxes[r-1] together.
Time Complexity: O(n^4). n = boxes.length. dfs takes O(n). memo is 3 dimensional, thus it takes O(n^4) totally.
Space: O(n^3).
AC Java:
1 class Solution {
2 public int removeBoxes(int[] boxes) {
3 int n = boxes.length;
4 int [][][] dp = new int[n][n][n];
5 return removeBoxesSub(boxes, 0, n-1, 0, dp);
6 }
7
8 private int removeBoxesSub(int [] boxes, int l, int r, int k, int [][][] dp){
9 if(l>r){
10 return 0;
11 }
12
13 while(l<r && boxes[r]==boxes[r-1]){
14 r--;
15 k++;
16 }
17
18 if(dp[l][r][k] > 0){
19 return dp[l][r][k];
20 }
21
22 int res = removeBoxesSub(boxes, l, r-1, 0, dp) + (k+1)*(k+1);
23 for(int i = l; i<r; i++){
24 if(boxes[i] == boxes[r]){
25 res = Math.max(res, removeBoxesSub(boxes, l, i, k+1, dp)+removeBoxesSub(boxes, i+1, r-1, 0, dp));
26 }
27 }
28
29 dp[l][r][k] = res;
30 return res;
31 }
32 }