980. 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. 1 <= grid.length * grid[0].length <= 20

题目大意 #

在二维网格 grid 上，有 4 种类型的方格：

• 1 表示起始方格。且只有一个起始方格。
• 2 表示结束方格，且只有一个结束方格。
• 0 表示我们可以走过的空方格。
• -1 表示我们无法跨越的障碍。

返回在四个方向（上、下、左、右）上行走时，从起始方格到结束方格的不同路径的数目，每一个无障碍方格都要通过一次。

解题思路 #

• 这一题也可以按照第 79 题的思路来做。题目要求输出地图中从起点到终点的路径条数。注意路径要求必须走满所有空白的格子。
• 唯一需要注意的一点是，空白的格子并不是最后走的总步数，总步数 = 空白格子数 + 1，因为要走到终点，走到终点也算一步。

代码 #

package leetcode

var dir = [][]int{
	{-1, 0},
	{0, 1},
	{1, 0},
	{0, -1},
}

func uniquePathsIII(grid [][]int) int {
	visited := make([][]bool, len(grid))
	for i := 0; i < len(visited); i++ {
		visited[i] = make([]bool, len(grid[0]))
	}
	res, empty, startx, starty, endx, endy, path := 0, 0, 0, 0, 0, 0, []int{}
	for i, v := range grid {
		for j, vv := range v {
			switch vv {
			case 0:
				empty++
			case 1:
				startx, starty = i, j
			case 2:
				endx, endy = i, j
			}
		}
	}
	findUniquePathIII(grid, visited, path, empty+1, startx, starty, endx, endy, &res) // 可走的步数要加一，因为终点格子也算一步，不然永远走不到终点！
	return res
}

func isInPath(board [][]int, x, y int) bool {
	return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
}

func findUniquePathIII(board [][]int, visited [][]bool, path []int, empty, startx, starty, endx, endy int, res *int) {
	if startx == endx && starty == endy {
		if empty == 0 {
			*res++
		}
		return
	}
	if board[startx][starty] >= 0 {
		visited[startx][starty] = true
		empty--
		path = append(path, startx)
		path = append(path, starty)
		for i := 0; i < 4; i++ {
			nx := startx + dir[i][0]
			ny := starty + dir[i][1]
			if isInPath(board, nx, ny) && !visited[nx][ny] {
				findUniquePathIII(board, visited, path, empty, nx, ny, endx, endy, res)
			}
		}
		visited[startx][starty] = false
		//empty++ 这里虽然可以还原这个变量值，但是赋值没有意义，干脆不写了
		path = path[:len(path)-2]
	}
	return
}