题目：原题链接（中等）
标签：图、深度优先搜索、回溯算法、动态规划
| 解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
|---|---|---|---|
| Ans 1 (Python) | O ( N 2 ) O(N^2) O(N2) | O ( N 2 ) O(N^2) O(N2) | 56ms (72.42%) |
| Ans 2 (Python) | |||
| Ans 3 (Python) |
解法一（回溯算法，可实现向四个方向走）：
class Solution:
def __init__(self):
self.ans = []
def pathWithObstacles(self, obstacleGrid: List[List[int]]) -> List[List[int]]:
if obstacleGrid[0][0] == 1 or obstacleGrid[-1][-1] == 1:
return []
height, width = len(obstacleGrid), len(obstacleGrid[0])
visited = {(0, 0)}
path = [(0, 0)]
def dfs():
x, y = path[-1]
# 检查是否已走到终点
if x == height - 1 and y == width - 1:
self.ans = list(path)
else:
# 尝试向下走
if x < height - 1 and obstacleGrid[x + 1][y] != 1 and (x + 1, y) not in visited:
path.append((x + 1, y))
visited.add((x + 1, y))
dfs()
path.pop()
# 尝试向右走
if y < width - 1 and obstacleGrid[x][y + 1] != 1 and (x, y + 1) not in visited:
path.append((x, y + 1))
visited.add((x, y + 1))
dfs()
path.pop()
# # 尝试向上走
# if x > 0 and obstacleGrid[x - 1][y] != 1 and (x - 1, y) not in visited:
# path.append((x - 1, y))
# visited.add((x - 1, y))
# dfs()
# path.pop()
#
# # 尝试向左走
# if y > 0 and obstacleGrid[x][y - 1] != 1 and (x, y - 1) not in visited:
# path.append((x, y - 1))
# visited.add((x, y - 1))
# dfs()
# path.pop()
dfs()
return self.ans
