Leetcode| 62. 不同路径、63. 不同路径 II Day39

62. Unique Paths

dp[i][j] ：表示从(0, 0)出发，到(i, j) 有dp[i][j]条不同的路径
因为只能向下或向右移动，所以dp[i][j] = dp[i][j - 1] + dp[i - 1][j]
dp[i][0]一定都是1，因为从(0, 0)的位置到(i, 0)的路径只有一条，dp[0][j]也同理

class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        dp = [[0] * n for _ in range(m)]
        for i in range(m):
            dp[i][0] = 1
        for j in range(n):
            dp[0][j] = 1

        for i in range(1, m):
            for j in range(1, n):
                dp[i][j] = dp[i][j - 1] + dp[i - 1][j]

        return dp[m - 1][n - 1]

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63. Unique Paths II

递推公式和62.不同路径一样，dp[i][j] = dp[i - 1][j] + dp[i][j - 1]。
因为有了障碍，(i, j)如果就是障碍的话应该就保持初始状态（初始状态为0）。

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        row, col = len(obstacleGrid), len(obstacleGrid[0])
        dp = [[0] * col for _ in range(row)]

        for i in range(row):
            if obstacleGrid[i][0] == 1:     # 障碍之后的为0
                break
            dp[i][0] = 1
        for j in range(col):
            if obstacleGrid[0][j] == 1:
                break
            dp[0][j] = 1

        for i in range(1, row):
            for j in range(1, col):
                if obstacleGrid[i][j] == 0:
                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1]

        return dp[row - 1][col - 1]