62.不同路径

• 确定dp数组及其下标的含义：dp[i][j]代表机器人到达i行j列时不同的路径数
• 确定递推公式：dp[i][j] = dp[i-1][j] + dp[i][j-1]
• dp数组的初始化：dp[0][j] = 1 dp[i][0] = 1
• 确定遍历顺序：从前向后
• 举例推导dp数组

class Solution:def uniquePaths(self, m: int, n: int) -> int:dp = [[1] * n for _ in range(m)]for i in range(1, m):for j in range(1, n):dp[i][j] = dp[i-1][j] + dp[i][j-1]return dp[-1][-1]

可以使用一维数组优化空间复杂度

class Solution:def uniquePaths(self, m: int, n: int) -> int:dp = [1] * nfor i in range(1, m):for j in range(1, n):dp[j] += dp[j-1]return dp[-1]

63. 不同路径 II

• 确定dp数组以及下标的含义：dp[i][j]代表到达第i行第j列的不同路径数
• 确定递推公式：if obstacleGrid[i][j] == 1: dp[i][j] = 0 else: dp[i][j] = dp[i-1][j] + dp[i][j-1]
• dp数组的初始化：第0行和第0列都需要初始化为1，遇到障碍物后后续都初始化为0
• 确定遍历顺序：从前向后遍历
• 举例推导dp数组

class Solution:def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:m = len(obstacleGrid)n = len(obstacleGrid[0])dp = [[0] * n for _ in range(m)]for i in range(m):if obstacleGrid[i][0] == 0:dp[i][0] = 1else:breakfor j in range(n):if obstacleGrid[0][j] == 0:dp[0][j] = 1else:breakfor i in range(1, m):for j in range(1, n):if obstacleGrid[i][j] == 1:dp[i][j] = 0else:dp[i][j] = dp[i - 1][j] + dp[i][j - 1]return dp[-1][-1]