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Unique Paths III

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Question

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)

Solution

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/*
@param: int[][] grid
@return: int
Algorithm: 常规DFS+backtracking
*/
int res, todo = 1, sx, sy, ex, ey, row, col;
public int uniquePathsIII(int[][] grid) {
    row = grid.length;
    col = grid[0].length;
    if (row == 0 || col == 0) {
        return 0;
    }
    res = 0;
    for (int i = 0; i < row; i++) {
        for (int j = 0; j < col; j++) {
            if (grid[i][j] == 0) {
                todo++;
            }else if (grid[i][j] == 1) {
                sx = i;
                sy = j;
            }else if (grid[i][j] == 2) {
                ex = i;
                ey = j;
            }
        }
    }
    dfs(sx, sy, grid);
    return res;
}

public void dfs(int x, int y, int[][] grid) {
    if (x < 0 || y < 0 || x >= row || y >= col || grid[x][y] < 0) {
        return;
    }
    if (x == ex && y == ey ) {
        if (todo == 0){
            res++;                
        }
        return;
    }
    todo--;
    grid[x][y] = -2;
    dfs(x+1,y,grid);
    dfs(x-1,y,grid);
    dfs(x,y+1,grid);
    dfs(x,y-1,grid);
    todo++;
    grid[x][y] = 0;
}