# 980. 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.

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(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size(), x0 = 0, y0 = 0, target = 1, res = 0;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 1) {
x0 = i; y0 = j;
} else if (grid[i][j] == 0) {
++target;
}
}
}
helper(grid, target, x0, y0, res);
return res;
}
void helper(vector<vector<int>>& grid, int& target, int i, int j, int& res) {
int m = grid.size(), n = grid[0].size();
if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] < 0) return;
if (grid[i][j] == 2) {
if (target == 0) ++res;
return;
}
grid[i][j] = -2;
--target;
helper(grid, target, i + 1, j, res);
helper(grid, target, i - 1, j, res);
helper(grid, target, i, j + 1, res);
helper(grid, target, i, j - 1, res);
grid[i][j] = 0;
++target;
}
};
``````

Github 同步地址:

https://github.com/grandyang/leetcode/issues/980

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

https://leetcode.com/problems/unique-paths-iii/discuss/221941/C%2B%2B-brute-force-DFS

https://leetcode.com/problems/unique-paths-iii/discuss/221946/JavaPython-Brute-Force-Backtracking

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