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package com.fishercoder.solutions;
/**
* 546. Remove Boxes
*
* 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)
Note: The number of boxes n would not exceed 100.
*/
public class _546 {
public static class Solution1 {
/**
* credit: https://leetcode.com/articles/remove-boxes/#approach-2-using-dp-with-memorizationaccepted
*
* For an entry in dp[l][r][k], l represents the starting index of the subarray,
* r represents the ending index of the subarray
* and k represents the number of elements similar to the rth element
* following it which can be combined to obtain the point information to be stored in dp[l][r][k].
*/
public int removeBoxes(int[] boxes) {
int[][][] dp = new int[100][100][100];
return calculatePoints(boxes, dp, 0, boxes.length - 1, 0);
}
public int calculatePoints(int[] boxes, int[][][] dp, int l, int r, int k) {
if (l > r) {
return 0;
}
if (dp[l][r][k] != 0) {
return dp[l][r][k];
}
while (r > l && boxes[r] == boxes[r - 1]) {
r--;
k++;
}
dp[l][r][k] = calculatePoints(boxes, dp, l, r - 1, 0) + (k + 1) * (k + 1);
for (int i = l; i < r; i++) {
if (boxes[i] == boxes[r]) {
dp[l][r][k] = Math.max(dp[l][r][k],
calculatePoints(boxes, dp, l, i, k + 1) + calculatePoints(boxes, dp, i + 1, r - 1, 0));
}
}
return dp[l][r][k];
}
}
}
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