Python Leetcode980不同路径III

不同路径III

看了某同学今天的华为机试题，发现其第三题就跟这题一样，于是拿出来做一做。

典型的回溯算法练习题，代码如下：

class Solution:
    def __init__(self):
        self.out = 0
    def uniquePathsIII(self, grid: List[List[int]]) -> int:
        dircetions = [(1, 0), (-1, 0), (0, 1), (0, -1)]
        m, n = len(grid), len(grid[0])
        res = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 0:
                    res += 1
                if grid[i][j] == 1:
                    start = (i, j)
        visited = {start}
        self.backsearch(m, n, grid, visited, start[0], start[1], dircetions, res)
        return self.out
    def backsearch(self, m, n, grid, visited, i, j, dircetions, res):
        visited.add((i, j))
        if grid[i][j] == 2 and len(visited) == res + 2:
            return True
        for a, b in dircetions:
            i_new, j_new = a + i, b + j
            if 0 <= i_new < m and 0 <= j_new < n and (i_new, j_new) not in visited and grid[i_new][j_new] != -1:
                if self.backsearch(m, n, grid, visited | {(i_new, j_new)}, i_new, j_new, dircetions, res):
                    self.out += 1
        return False

别人的代码（可以打印路径）：

class Solution:
    def uniquePathsIII(self, grid: List[List[int]]) -> int:
        m, n=len(grid),len(grid[0])
        obstacle=set()
        for i in range(m):
            for j in range(n):
                if grid[i][j]==1: 
                    start=(i,j)
                elif grid[i][j]==2:
                    end=(i,j)
                elif grid[i][j]==-1:
                    obstacle.add((i,j))
        total_step=n*m-len(obstacle)
        visited=set([start]) 
        res=[0]    
        direcs=[[-1,0],[1,0],[0,1],[0,-1]]
        def dfs(pos,path):
            if len(path)==total_step and pos==end:
                print(path)
                res[0]+=1
                return
            for dx, dy in direcs:
                x, y=pos[0]+dx, pos[1]+dy
                if 0<=x<m and 0<=y<n and (x,y) not in visited\
                    and (x,y) not in obstacle:
                    visited.add((x,y))
                    dfs((x,y), path+[(x,y)])
                    visited.remove((x,y))
        dfs(start,[start])
        return res[0]

直接在原数组上修改已访问的方格，将其设置为-1.

class Solution:
    def uniquePathsIII(self, grid):
        start, end, p = None, None, 1
        row, col = len(grid), len(grid[0])
        for i in range(row):
            for j in range(col):
                if grid[i][j] == 1:
                    start = i, j
                elif grid[i][j] == 2:
                    end = i, j
                elif grid[i][j] == 0:
                    p += 1   
        def dfs(x, y, p):
            if not (0 <= x < row and 0 <= y < col and grid[x][y] >= 0):
                return 0
            if end == (x, y) and p == 0:
                return 1
            grid[x][y] = -1
            res = dfs(x+1, y, p-1) + dfs(x, y+1, p-1) + dfs(x-1, y, p-1) + dfs(x, y-1, p-1)
            grid[x][y] = 0
            return res
        return dfs(start[0], start[1], p)