1219. Path with Maximum Gold

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:

• Every time you are located in a cell you will collect all the gold in that cell.

• From your position, you can walk one step to the left, right, up, or down.

• You can't visit the same cell more than once.

• Never visit a cell with 0 gold.

• You can start and stop collecting gold from any position in the grid that has some 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

• There are at most 25 cells containing gold.

解題

func getMaximumGold(grid [][]int) int {
    ans := 0 

    var backtrack func(int, int, int)
    backtrack = func(x int, y int, now int) {
        // invalid index
        if x >= len(grid) || y >= len(grid[0]) || x < 0 || y < 0 {
            return 
        }
        
        // 小於0代表走過了 等於0代表不能走
        if grid[x][y] < 0 || grid[x][y] == 0 {
            return
        }
        
        // 舊的值加上目前這格的值
        newValue := now+grid[x][y]
        if newValue > ans {
            ans = newValue
        }

        origin := grid[x][y] //因為之後要backtrack，所以改值之前要把原本的存起來
        grid[x][y] = -1
        backtrack(x+1, y, newValue)
        backtrack(x-1, y, newValue)
        backtrack(x, y+1, newValue)
        backtrack(x, y-1, newValue)
        grid[x][y] = origin
    }

    for i:=0;i<len(grid); i++ {
        for j:=0; j<len(grid[0]); j++ {
            backtrack(i, j, 0)
        }
    }

    return ans
}