代码随想录算法训练营day42 | 62.不同路径、63. 不同路径 II

62.不同路径

• 确定dp数组及其下标的含义：dpij代表机器人到达i行j列时不同的路径数
• 确定递推公式：dpij = dpi-1j + dpij-1
• dp数组的初始化：dp0j = 1 dpi0 = 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] * n
        for 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：dpij代表到达第i行第j列的不同路径数
• 确定递推公式：if obstacleGridij == 1: dpij = 0 else: dpij = dpi-1j + dpij-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] = 1
            else:
                break
        for j in range(n):
            if obstacleGrid[0][j] == 0:
                dp[0][j] = 1
            else:
                break
        for i in range(1, m):
            for j in range(1, n):
                if obstacleGrid[i][j] == 1:
                    dp[i][j] = 0
                else:
                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
        return dp[-1][-1]