LeetCode 980. Unique Paths III

原题链接在这里：https://leetcode.com/problems/unique-paths-iii/

题目：

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

题解：

The DFS states need current coordinate, current count of 0 position, target count of 0 position, and visited grid.

If current coordinate is out of bound, or grid value is -1 or it is visited before, simply return.

If it is current coordinate is target coordinate, if current 0 count == target count, we find a path. Whether or not this is a path, we need to return here.

It its value is 0, accumlate 0 count.

Mark this position as visited and for 4 dirs, continue DFS.

Backtracking needs to reset visited as false at this coordinate.

Time Complexity: exponential.

Space: O(m*n). m = grid.length. n = grid[0].length.

AC Java:

class Solution {
    int[][] dirs = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
    int res = 0;
    public int uniquePathsIII(int[][] grid) {
        if(grid == null || grid.length == 0 || grid[0].length == 0){
            return 0;
        }

        int m = grid.length;
        int n = grid[0].length;
        int startX = -1;
        int startY = -1;
        int zeroCount = 0;
        
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(grid[i][j] == 1){
                    startX = i;
                    startY = j;
                }else if(grid[i][j] == 0){
                    zeroCount++;
                }
            }
        }

        dfs(grid, startX, startY, m, n, 0, zeroCount, new boolean[m][n]);
        return res;
    }

    private void dfs(int[][] grid, int i, int j, int m, int n, int count, int targetCount, boolean[][] visited){
        if(i < 0 || i >= m || j < 0 || j >= n || visited[i][j] || grid[i][j] == -1){
            return;
        }

        if(grid[i][j] == 2){
            if(count == targetCount){
                res++;
            }

            return;
        }

        if(grid[i][j] == 0){
            count++;
        }

        visited[i][j] = true;
        for(int [] dir : dirs){
            int dx = i + dir[0];
            int dy = j + dir[1];
            dfs(grid, dx, dy, m, n, count, targetCount, visited);
        }
        visited[i][j] = false;
    }
}

类似Sudoku Solver, Unique Paths, Unique Paths II.