标签：rect can visit math min ted cat imu output
In a gold mine grid of size m * n, each cell in this mine has an integer representing the amount of gold in that cell, 0 if it is empty.
Return the maximum amount of gold you can collect under the conditions:
0 gold.
Example 1:
Input: grid = [[0,6,0],[5,8,7],[0,9,0]] Output: 24 Explanation: [[0,6,0], [5,8,7], [0,9,0]] Path to get the maximum gold, 9 -> 8 -> 7.
Example 2:
Input: grid = [[1,0,7],[2,0,6],[3,4,5],[0,3,0],[9,0,20]] Output: 28 Explanation: [[1,0,7], [2,0,6], [3,4,5], [0,3,0], [9,0,20]] Path to get the maximum gold, 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7.
class Solution { int gRow; int gCol; int[][] directions = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}}; public int getMaximumGold(int[][] grid) { gRow = grid.length; gCol = grid[0].length; int res = 0; boolean[][] visited = new boolean[gRow][gCol]; for (int i = 0; i < gRow; i++) { for (int j = 0; j < gCol; j++) { res = Math.max(res, dfs(grid, i, j, visited)); } } return res; } private int dfs(int[][] grid, int row, int col, boolean[][] visited) { if (row < 0 || row >= gRow || col < 0 || col >= gCol || grid[row][col] == 0 || visited[row][col]) { return 0; } int curValue = grid[row][col]; int nxtMax = 0; // mark as visited visited[row][col] = true; for (int[] direction: directions) { nxtMax = Math.max(nxtMax, dfs(grid, row + direction[0], col + direction[1], visited)); } visited[row][col] = false; return curValue + nxtMax; } }
[LC] 1219. Path with Maximum Gold
标签：rect can visit math min ted cat imu output
原文地址：https://www.cnblogs.com/xuanlu/p/12791216.html
