[Swift]LeetCode1219. 黄金矿工 | Path with Maximum Gold

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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:

• 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:

• 1 <= grid.length, grid[i].length <= 15
• 0 <= grid[i][j] <= 100
• There are at most 25 cells containing gold.

---

你要开发一座金矿，地质勘测学家已经探明了这座金矿中的资源分布，并用大小为 m * n 的网格 grid 进行了标注。每个单元格中的整数就表示这一单元格中的黄金数量；如果该单元格是空的，那么就是 0。

为了使收益最大化，矿工需要按以下规则来开采黄金：

• 每当矿工进入一个单元，就会收集该单元格中的所有黄金。
• 矿工每次可以从当前位置向上下左右四个方向走。
• 每个单元格只能被开采（进入）一次。
• 不得开采（进入）黄金数目为 0 的单元格。
• 矿工可以从网格中 任意一个 有黄金的单元格出发或者是停止。

示例 1：

输入：grid = [[0,6,0],[5,8,7],[0,9,0]]
输出：24
解释：
[[0,6,0],
 [5,8,7],
 [0,9,0]]
一种收集最多黄金的路线是：9 -> 8 -> 7。

示例 2：

输入：grid = [[1,0,7],[2,0,6],[3,4,5],[0,3,0],[9,0,20]]
输出：28
解释：
[[1,0,7],
 [2,0,6],
 [3,4,5],
 [0,3,0],
 [9,0,20]]
一种收集最多黄金的路线是：1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7。

提示：

• 1 <= grid.length, grid[i].length <= 15
• 0 <= grid[i][j] <= 100
• 最多 25 个单元格中有黄金。

---

Runtime: 60 ms

Memory Usage: 21.1 MB

class Solution {
    let dir:[[Int]] = [[-1,0],[0,-1],[0,1],[1,0]]
    func getMaximumGold(_ grid: [[Int]]) -> Int {
        let m:Int = grid.count
        let n:Int = grid[0].count
        var res:Int = 0
        for i in 0..<m
        {
            for j in 0..<n
            {
                if grid[i][j] == 0 {continue}
                let num:Int = countneighbor(grid,i,j)
                if !((i==0&&j==0)||(i==m-1&&j==0)||(i==m-1&&j==n-1)||(i==0&&j==n-1))&&num>1 {continue}
                var seen:[[Bool]] = [[Bool]](repeating: [Bool](repeating: false, count: n), count: m)
                let cur:Int = dfs(grid,i,j,&seen)
                res = max(cur,res)
            }
        }
        return res
    }
    
    func dfs(_ grid: [[Int]],_ i:Int,_ j:Int,_ seen:inout [[Bool]]) -> Int
    {
        let m:Int = grid.count
        let n:Int = grid[0].count
        if i>=m||i<0||j>=n||j<0||seen[i][j]||grid[i][j] == 0 {return 0}
        seen[i][j] = true
        let l:Int = dfs(grid,i-1,j,&seen)
        let r:Int = dfs(grid,i+1,j,&seen)
        let u:Int = dfs(grid,i,j - 1,&seen)
        let d:Int = dfs(grid,i,j + 1,&seen)
        seen[i][j] = false
        return grid[i][j] + max(l,max(r,max(u,d)))
    }
    
    func countneighbor(_ grid: [[Int]],_ i:Int,_ j:Int) -> Int
    {
        var count:Int = 0
        let m:Int = grid.count
        let n:Int = grid[0].count
        for d in dir
        {
            let x:Int = d[0]+i
            let y:Int = d[1]+j
            if x>=0 && x<m && y>=0 && y<n && grid[x][y] != 0
            {
                count += 1
            }
        }
        return count
    }
}