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:

  1. Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]]
  2. Output: 2
  3. Explanation: We have the following two paths: 
  4. 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2)
  5. 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:

  1. Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]]
  2. Output: 4
  3. Explanation: We have the following four paths: 
  4. 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)
  5. 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)
  6. 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)
  7. 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:

  1. Input: [[0,1],[2,0]]
  2. Output: 0
  3. Explanation: 
  4. There is no path that walks over every empty square exactly once.
  5. 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

  1. class Solution {
  2.     private int count;
  3.     private int m, n;
  4.     private int[] iPlus = { -1, 0, 1, 0 };
  5.     private int[] jPlus = { 0, 1, 0, -1 };
  7.     public int uniquePathsIII(int[][] grid) {
  8.         m = grid.length;
  9.         n = grid[0].length;
  10.         int x = 0, y = 0;
  11.         int left = 0;
  13.         for (int i = 0; i < m; i++) {
  14.             for (int j = 0; j < n; j++) {
  15.                 if (grid[i][j] >= 0) {
  16.                     left++;
  17.                 }
  19.                 if (grid[i][j] == 1) {
  20.                     x = i;
  21.                     y = j;
  22.                 }
  23.             }
  24.         }
  26.         dfs(grid, x, y, left, new boolean[m][n]);
  27.         return count;
  28.     }
  30.     private void dfs(int[][] grid, int i, int j, int left, boolean visited[][]) {
  31.         if (grid[i][j] == 2 && left == 1) {
  32.             count++;
  33.             return;
  34.         }
  36.         visited[i][j] = true;
  37.         for (int x = 0; x < 4; x++) {
  38.             int nextI = i + iPlus[x];
  39.             int nextJ = j + jPlus[x];
  40.             if (isValid(grid, nextI, nextJ) && !visited[nextI][nextJ]) {
  41.                 dfs(grid, nextI, nextJ, left - 1, visited);
  42.             }
  43.         }
  45.         visited[i][j] = false;
  46.     }
  48.     private boolean isValid(int[][] grid, int i, int j) {
  49.         return i >= 0 && i < m && j >= 0 && j < n && grid[i][j] != -1;
  50.     }
  51. }