标签：present pos bool set lse ack title tco square
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 <= grid.length * grid[0].length <= 20题解：
The DFS states need current coordinate, target coordinate, current count of 0 position, target count of 0 position, and visited grid.
If current coordinate is out of bound, or its 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 we 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:
1 class Solution { 2 int pathCount = 0; 3 int [][] dirs = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}}; 4 5 public int uniquePathsIII(int[][] grid) { 6 if(grid == null || grid.length == 0){ 7 return 0; 8 } 9 10 int m = grid.length; 11 int n = grid[0].length; 12 int startX = -1; 13 int startY = -1; 14 int endX = -1; 15 int endY = -1; 16 int zeroCount = 0; 17 18 for(int i = 0; i<m; i++){ 19 for(int j = 0; j<n; j++){ 20 if(grid[i][j] == 1){ 21 startX = i; 22 startY = j; 23 }else if(grid[i][j] == 2){ 24 endX = i; 25 endY = j; 26 }else if(grid[i][j] == 0){ 27 zeroCount++; 28 } 29 } 30 } 31 32 dfs(grid, startX, startY, endX, endY, 0, zeroCount, new boolean[m][n]); 33 return pathCount; 34 } 35 36 private void dfs(int [][] grid, int i, int j, int endX, int endY, int count, int targetCount, boolean [][] visited){ 37 if(i < 0 || i >= grid.length || j < 0 || j>= grid[0].length || grid[i][j] == -1 || visited[i][j]){ 38 return; 39 } 40 41 if(grid[i][j] == 2){ 42 if(count == targetCount){ 43 pathCount++; 44 } 45 46 return; 47 } 48 49 if(grid[i][j] == 0){ 50 count++; 51 } 52 53 visited[i][j] = true; 54 for(int [] dir : dirs){ 55 int dx = i + dir[0]; 56 int dy = j + dir[1]; 57 dfs(grid, dx, dy, endX, endY, count, targetCount, visited); 58 } 59 60 visited[i][j] = false; 61 } 62 }
类似Sudoku Solver, Unique Paths, Unique Paths II.
标签：present pos bool set lse ack title tco square
原文地址：https://www.cnblogs.com/Dylan-Java-NYC/p/12047339.html
