[Swift]LeetCode980. 不同路径 III | Unique Paths III

原文地址：https://www.cnblogs.com/strengthen/p/10295241.html 

On a 2-dimensional grid, there are 4 types of squares:

• 1 represents the starting square.  There is exactly one starting square.
• 2 represents the ending square.  There is exactly one ending square.
• 0 represents empty squares we can walk over.
• -1 represents obstacles that we cannot walk over.

Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once. 

Example 1:

Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]]
Output: 2
Explanation: We have the following two paths: 
1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2)
2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2)

Example 2:

Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]]
Output: 4
Explanation: We have the following four paths: 
1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3)
2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3)
3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3)
4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3)

Example 3:

Input: [[0,1],[2,0]]
Output: 0
Explanation: 
There is no path that walks over every empty square exactly once.
Note that the starting and ending square can be anywhere in the grid. 

Note:

1. 1 <= grid.length * grid[0].length <= 20

---

在二维网格 grid 上，有 4 种类型的方格：

• 1 表示起始方格。且只有一个起始方格。
• 2 表示结束方格，且只有一个结束方格。
• 0 表示我们可以走过的空方格。
• -1 表示我们无法跨越的障碍。

返回在四个方向（上、下、左、右）上行走时，从起始方格到结束方格的不同路径的数目，每一个无障碍方格都要通过一次。 

示例 1：

输入：[[1,0,0,0],[0,0,0,0],[0,0,2,-1]]
输出：2
解释：我们有以下两条路径：
1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2)
2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2)

示例 2：

输入：[[1,0,0,0],[0,0,0,0],[0,0,0,2]]
输出：4
解释：我们有以下四条路径： 
1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3)
2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3)
3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3)
4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3)

示例 3：

输入：[[0,1],[2,0]]
输出：0
解释：
没有一条路能完全穿过每一个空的方格一次。
请注意，起始和结束方格可以位于网格中的任意位置。 

提示：

1. 1 <= grid.length * grid[0].length <= 20

---

32ms

class Solution {
    var zero:Int = 0
    var ans:Int = 0
    func uniquePathsIII(_ grid: [[Int]]) -> Int {
        var start1:Int = 0
        var start2:Int = 0
        for i in 0..<grid.count
        {
            for j in 0..<grid[0].count
            {
                if grid[i][j] == 0
                {
                    zero += 1
                }
                if grid[i][j] == 1
                {
                    start1 = i
                    start2 = j
                }
            }            
        }
        var visited:[[Int]] = [[Int]](repeating:[Int](repeating:0,count:grid[0].count),count:grid.count)
        dfs(grid, start1, start2, visited, 0)
        return ans
    }
    
    func dfs(_ grid: [[Int]],_ i:Int,_ j:Int,_ visited: [[Int]],_ count:Int)
    {
        var  visited = visited
        if i < 0 || i >= grid.count || j < 0 || j >= grid[0].count || visited[i][j] == 1 || grid[i][j] == -1
        {
            return
        }
        if grid[i][j] == 2
        {
            if count == zero + 1
            {
                ans += 1
            }
            return
        }
        visited[i][j] = 1
        dfs(grid, i+1, j, visited, count+1)
        dfs(grid, i-1, j, visited, count+1)
        dfs(grid, i, j-1, visited, count+1)
        dfs(grid, i, j+1, visited, count+1)
        visited[i][j] = 0
    }
}