Medium
In a gold mine grid
of size m x 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.
Constraints:
m == grid.length
n == grid[i].length
1 <= m, n <= 15
0 <= grid[i][j] <= 100
public class Solution {
private int maxGold = 0;
public int getMaximumGold(int[][] grid) {
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (grid[i][j] != 0) {
int g = grid[i][j];
grid[i][j] = 0;
gold(grid, i, j, g);
grid[i][j] = g;
}
}
}
return maxGold;
}
private void gold(int[][] grid, int row, int col, int gold) {
if (gold > maxGold) {
maxGold = gold;
}
if (row > 0 && grid[row - 1][col] != 0) {
int currGold = grid[row - 1][col];
grid[row - 1][col] = 0;
gold(grid, row - 1, col, gold + currGold);
grid[row - 1][col] = currGold;
}
if (col > 0 && grid[row][col - 1] != 0) {
int currGold = grid[row][col - 1];
grid[row][col - 1] = 0;
gold(grid, row, col - 1, gold + currGold);
grid[row][col - 1] = currGold;
}
if (row < grid.length - 1 && grid[row + 1][col] != 0) {
// flag=false;
int currGold = grid[row + 1][col];
grid[row + 1][col] = 0;
gold(grid, row + 1, col, gold + currGold);
grid[row + 1][col] = currGold;
}
if (col < grid[0].length - 1 && grid[row][col + 1] != 0) {
int currGold = grid[row][col + 1];
grid[row][col + 1] = 0;
gold(grid, row, col + 1, gold + currGold);
grid[row][col + 1] = currGold;
}
}
}