代码随想录算法训练营day39 | 62.不同路径、63. 不同路径 II
62.不同路径
- dp数组以及下标的含义：dp[i][j]代表到达第i行第j列有多少条不同的路径
- 递推公式：dp[i][j] = dp[i][j-1] + dp[i-1][j]
- dp数组初始化：dp[0][:] = 1 dp[:][0] = 1
- 遍历顺序：从前往后遍历
- 举例推导dp数组：
按照这种方法自己做出来了，有点进步
class Solution:def uniquePaths(self, m: int, n: int) -> int:dp = [[0] * n for _ in range(m)]for j in range(n):dp[0][j] = 1for i in range(m):dp[i][0] = 1for i in range(1, m):for j in range(1, n):dp[i][j] = dp[i][j - 1] + dp[i - 1][j]return dp[-1][-1]63. 不同路径 II
遇到障碍的时候dp赋值为0，要注意在初始化的时候，只要遇到障碍，后面的数值都置为0，因为不可能到达了
- dp数组以及下标的含义：dp[i][j]代表到达第i行第j列有多少条不同的路径
- 递推公式：dp[i][j] = dp[i][j-1] + dp[i-1][j]
- dp数组初始化：dp[0][:] = 1 dp[:][0] = 1
- 遍历顺序：从前往后遍历
- 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:breakelse:dp[i][0] = 1for j in range(n):if obstacleGrid[0][j] != 0:breakelse:dp[0][j] = 1for 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][j-1] + dp[i-1][j]return dp[-1][-1]注意：
- 可以直接在obstacleGrid上更改，节约空间
- 如果起点或终点有障碍物，直接返回0 即if obstacleGrid[m - 1][n - 1] == 1 or obstacleGrid[0][0] == 1: return 0