LeetCode Unique Paths

LeetCode Unique Paths
Solving LeetCode with Unique Paths

Original question

The number of different paths from the start point to the end point of a robot can only go to the right or down.

Note:

The maximum grid size is 100*100.

Example:

Input: m = 3, n = 7

Output: 28

Solutions

A common primary school student's Mathematical Olympiad problem can be solved by means of permutation and combination. A total of steps (S-1) + (n-1) should be taken, m-1) steps down, (n-1) to the right, and there is a formulamCn = n!/m!(n-m)!, You can use the following code to solve the problem:

import mathclass Solution(object):    def uniquePaths(self, m, n):        """        :type m: int        :type n: int        :rtype: int        """        m -= 1        n -= 1        return math.factorial(m+n) / (math.factorial(n) * math.factorial(m))

Of course, a more common approach is dynamic planning. To reach a grid, you only need to come over from the grid above it or on the left. Recursive relationship:dp[i][j]=dp[i-1][j]+dp[i][j-1]. The initialization condition is that there is only one path on the left and the top, and all grids are initialized to 1 during initialization.

AC Source Code

class Solution(object):    def uniquePaths(self, m, n):        """        :type m: int        :type n: int        :rtype: int        """        dp = [[1 for __ in range(n)] for __ in range(m)]        for i in range(1, n):            for j in range(1, m):                dp[j][i] = dp[j - 1][i] + dp[j][i - 1]        return dp[m - 1][n - 1]if __name__ == "__main__":    assert Solution().uniquePaths(3, 7) == 28