leetcode 546. Remove Boxes

Solution 讲解： leetcode 546. Remove Boxes

很巧妙的分析思路：T（i, j, k) 来表示 k个相同前缀，i-j 之间的最大值

自己写的自底向上的 Bottom-Top

public class Solution {public int removeBoxes(int[] boxes) {int len = boxes.length;if(len==0) return 0;int[][][] dp = new int[len][len][len+1];for(int inter=0;inter<len;inter++){for(int i=0;i+inter<len;i++){for(int k=0;k<len;k++){int j = i+inter; //j>=i// if j<i  => 0dp[i][j][k] = 0;if(i==j)  {dp[i][j][k]=(k+1)*(k+1);continue;}if(dp[i+1][j][0]+(k+1)*(k+1)>dp[i][j][k]){dp[i][j][k] = dp[i+1][j][0]+(k+1)*(k+1);}for(int h=i+1;h<=j;h++){if(boxes[h]==boxes[i]){if(dp[i+1][h-1][0] + dp[h][j][k+1]>dp[i][j][k]){dp[i][j][k] = dp[i+1][h-1][0] + dp[h][j][k+1];}}}}}}return dp[0][len-1][0];}
}

这一题很经典的dp题目，Solution给出如下的写法，也是dp常见的套路：

Top-Bottom:

public int removeBoxes(int[] boxes) {int n = boxes.length;int[][][] dp = new int[n][n][n];return removeBoxesSub(boxes, 0, n - 1, 0, dp);
}private int removeBoxesSub(int[] boxes, int i, int j, int k, int[][][] dp) {if (i > j) return 0;if (dp[i][j][k] > 0) return dp[i][j][k];int res = (k + 1) * (k + 1) + removeBoxesSub(boxes, i + 1, j, 0, dp);for (int m = i + 1; m <= j; m++) {if (boxes[i] == boxes[m]) {res = Math.max(res, removeBoxesSub(boxes, i + 1, m - 1, 0, dp) + removeBoxesSub(boxes, m, j, k + 1, dp));}}dp[i][j][k] = res;return res;
}

Bottom-Top:

public int removeBoxes(int[] boxes) {int n = boxes.length;int[][][] dp = new int[n][n][n];for (int j = 0; j < n; j++) {for (int k = 0; k <= j; k++) {dp[j][j][k] = (k + 1) * (k + 1);}}for (int l = 1; l < n; l++) {for (int j = l; j < n; j++) {int i = j - l;for (int k = 0; k <= i; k++) {int res = (k + 1) * (k + 1) + dp[i + 1][j][0];for (int m = i + 1; m <= j; m++) {if (boxes[m] == boxes[i]) {res = Math.max(res, dp[i + 1][m - 1][0] + dp[m][j][k + 1]);}}dp[i][j][k] = res;}}}return (n == 0 ? 0 : dp[0][n - 1][0]);
}

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