LeetCode 741. Cherry Pickup

原题链接在这里：https://leetcode.com/problems/cherry-pickup/

题目：

In a N x N grid representing a field of cherries, each cell is one of three possible integers.

• 0 means the cell is empty, so you can pass through;
• 1 means the cell contains a cherry, that you can pick up and pass through;
• -1 means the cell contains a thorn that blocks your way.

Your task is to collect maximum number of cherries possible by following the rules below:

• Starting at the position (0, 0) and reaching (N-1, N-1) by moving right or down through valid path cells (cells with value 0 or 1);
• After reaching (N-1, N-1), returning to (0, 0) by moving left or up through valid path cells;
• When passing through a path cell containing a cherry, you pick it up and the cell becomes an empty cell (0);
• If there is no valid path between (0, 0) and (N-1, N-1), then no cherries can be collected.

Example 1:

Input: grid =
[[0, 1, -1],
 [1, 0, -1],
 [1, 1,  1]]
Output: 5
Explanation:
The player started at (0, 0) and went down, down, right right to reach (2, 2).
4 cherries were picked up during this single trip, and the matrix becomes [[0,1,-1],[0,0,-1],[0,0,0]].
Then, the player went left, up, up, left to return home, picking up one more cherry.
The total number of cherries picked up is 5, and this is the maximum possible.

Note:

• grid is an N by N 2D array, with 1 <= N <= 50.
• Each grid[i][j] is an integer in the set {-1, 0, 1}.
• It is guaranteed that grid[0][0] and grid[N-1][N-1] are not -1.

题解：

It could be understanding as two people collecting cheeries from (n-1, n-1) to (0, 0).

Two people cooridinates are (x1, y1), (x2, y2).

dfs(x1, y1, x2, y2) returns maximum cheeries collected from two cooridinates to (0, 0).

Thus max(x1, y1, x2, y2) = grid[x1][y1] + grid[x2][y2] + max(dfs(x1-1, y1, x2-1, y2), dfs(x1, y1-1, x2, y2-1), dfs(x1-1, y1, x2, y2-1), dfs(x1, y1-1, x2-1, y2)).

First person could move from top, or left. Second person could do the same. Totally 4 combinations.

And of cource, if x1==y1, which means both people are on the same grid, its cheery can't be collected twice.

y2 = x1+y1-x2. since both of them have same total steps.

Time Complexity: O(n^3).

Space: O(n^3).

AC Java:

 class Solution {
     int [][][] dp;
     int n;
     public int cherryPickup(int[][] grid) {
         if(grid == null || grid.length == 0 || grid[0].length == 0){
             return 0;
         }

         n = grid.length;
         dp = new int[n][n][n];
         for(int i = 0; i<n; i++){
             for(int j = 0; j<n; j++){
                 Arrays.fill(dp[i][j], Integer.MIN_VALUE);
             }
         }

         return Math.max(0, dfs(grid, n-1, n-1, n-1));
     }

     private int dfs(int [][] grid, int x1, int y1, int x2){
         int y2 = x1+y1-x2;
         if(x1<0 || y1<0 || x2<0 || y2<0){
             return -1;
         }

         if(grid[x1][y1]<0 || grid[x2][y2]<0){
             return -1;
         }

         if(dp[x1][y1][x2] != Integer.MIN_VALUE){
             return dp[x1][y1][x2];
         }

         if(x1==0 && y1==0){
             dp[0][0][0] = grid[0][0];
             return grid[0][0];
         }

         int res = Math.max(Math.max(dfs(grid, x1-1, y1, x2-1), dfs(grid, x1, y1-1, x2)), Math.max(dfs(grid, x1-1, y1, x2), dfs(grid, x1, y1-1, x2-1)));
         if(res < 0){
             dp[x1][y1][x2] = -1;
             return -1;
         }

         res += grid[x1][y1];
         if(x1 != x2){
             res += grid[x2][y2];
         }

         dp[x1][y1][x2] = res;
         return res;
     }
 }