0980. Unique Paths III (H)

Unique Paths III (H)

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

---

题意

给定一个二维数组，1表示起点，2表示终点，0表示可通行，-1表示不可通行。找到一条从起点到终点的路径，使其能通过所有可通行的点，统计这样的路径的个数。

思路

直接暴力回溯就能AC。

---

代码实现

Java

class Solution {
    private int count;
    private int m, n;
    private int[] iPlus = { -1, 0, 1, 0 };
    private int[] jPlus = { 0, 1, 0, -1 };

    public int uniquePathsIII(int[][] grid) {
        m = grid.length;
        n = grid[0].length;
        int x = 0, y = 0;
        int left = 0;

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] >= 0) {
                    left++;
                }

                if (grid[i][j] == 1) {
                    x = i;
                    y = j;
                }
            }
        }

        dfs(grid, x, y, left, new boolean[m][n]);
        return count;
    }

    private void dfs(int[][] grid, int i, int j, int left, boolean visited[][]) {
        if (grid[i][j] == 2 && left == 1) {
            count++;
            return;
        }

        visited[i][j] = true;
        for (int x = 0; x < 4; x++) {
            int nextI = i + iPlus[x];
            int nextJ = j + jPlus[x];
            if (isValid(grid, nextI, nextJ) && !visited[nextI][nextJ]) {
                dfs(grid, nextI, nextJ, left - 1, visited);
            }
        }

        visited[i][j] = false;
    }

    private boolean isValid(int[][] grid, int i, int j) {
        return i >= 0 && i < m && j >= 0 && j < n && grid[i][j] != -1;
    }
}