标签：Paths square int res 980 dfs length grid Unique

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

class Solution {
    public int uniquePathsIII(int[][] grid) {
        int n = grid.length * grid[0].length;
        int[] res = new int[1];
        for(int i = 0; i < grid.length; i++) {
            for(int j = 0; j < grid[0].length; j++) {
                if(grid[i][j] == 1) dfs(grid, i, j, res, 0, n);
            }
        }
        return res[0];
    }
    public void dfs(int[][] grid, int i, int j, int[] res, int count, int n) {
        int cur = grid[i][j];
        if(cur == 2 && count == n - 1) {
            res[0]++;
            return;
        }
        if(cur == -1) return;
        
        grid[i][j] = -1;
        if(i > 0) dfs(grid, i - 1, j, res, count + 1, n);
        if(i < grid.length - 1) dfs(grid, i + 1, j, res, count + 1, n);
        if(j > 0) dfs(grid, i, j - 1, res, count + 1, n);
        if(j < grid[0].length - 1) dfs(grid, i, j + 1, res, count + 1, n);
        grid[i][j] = cur;
    }
}

38/39?尼玛

class Solution {
    int res = 0, empty = 1, sx, sy;
    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] == -1)
            return;
        if (grid[x][y] == 2) {
            if (empty == 0) res++;
            return;
        }
        grid[x][y] = -1;
        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;
        empty++;
    }
}

答案来自lee哥，没啥区别我觉得？

标签：Paths,square,int,res,980,dfs,length,grid,Unique
来源： https://www.cnblogs.com/wentiliangkaihua/p/13722207.html

本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享；
2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关；
3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关；
4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除；
5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。