LeetCode62_ 不同路径

1. 题目

2. 题解

2.1 解法1 ：动态规划

# 62
class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        dp = [[0] * n for _ in range(m)]
        dp[0][0] = 1
        for i in range(m):
            for j in range(n):
                if 0 <= i - 1 < m:
                    dp[i][j] = dp[i][j] + dp[i - 1][j]
                if 0 <= j - 1 < n:
                    dp[i][j] = dp[i][j] + dp[i][j - 1]
        return dp[m -1][n -1]


if __name__ == "__main__":
    s = Solution()
    a = s.uniquePaths(3,7)
    print(a)

2.2 解法2： 动态规划

class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        # 给边缘化的值赋值为1
        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[m - 1][n - 1]

2.3 解法3 ：解法2空间复杂度优化

class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        # 给边缘化的值赋值为1
        dp = [1] * n
        for _ in range(1, m):
            for j in range(1, n):
                dp[j] += dp[j - 1]
        return dp[n - 1]