980. Unique Paths III (Hard).md

980. Unique Paths III (Hard)

https://leetcode.com/problems/unique-paths-iii/

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.

代码

• Backtrack

class Solution {
    int res = 0, empty = 1, sx, sy, ex, ey;
    public int uniquePathsIII(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 0) 
                    empty++;
                else if (grid[i][j] == 1) {
                    sx = i;
                    sy = j;
                }
            }
        }
        dfs(grid, sx, sy);
        return res;
    }

    public void dfs(int[][] grid, int x, int y) {
        if (x < 0 || x >= grid.length || y < 0 || y >= grid[0].length || grid[x][y] < 0)
            return;
        if (grid[x][y] == 2) {
            if (empty == 0) res++;
            return;
        }
        grid[x][y] = -666;	// change to visited
        empty--;
        dfs(grid, x + 1, y);
        dfs(grid, x - 1, y);
        dfs(grid, x, y + 1);
        dfs(grid, x, y - 1);
        grid[x][y] = 0;		// reset to unvisted
        empty++;
    }
}

• Python

  class Solution(object):
      def uniquePathsIII(self, A):
          """
          :type grid: List[List[int]]
          :rtype: int
          """
          '''
          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.
          '''
          
          self.res = 0
          m, n = len(A), len(A[0])
          empty = 1
          for i in range(m):
              for j in range(n):
                  if A[i][j] == 1:
                      sx, sy = (i, j)
                  elif A[i][j] == 0:
                      empty += 1
          
          def dfs(x, y, empty):
              if not (0 <= x < m and 0 <= y < n and A[x][y] >= 0):
                  return 
              if A[x][y] == 2:
                  # destiny
                  self.res += empty == 0
                  return
              
              A[x][y] = -666 # change
              dfs(x+1, y, empty-1)
              dfs(x-1, y, empty-1)
              dfs(x, y+1, empty-1)
              dfs(x, y-1, empty-1)
              A[x][y] = 0 # recover
              
          dfs(sx, sy, empty)
          return self.res