problem

A robot is located at the top-left corner of a m x n grid (marked ‘Start’ in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish’ in the diagram below).

How many possible unique paths are there?

solution

这个问题是爬楼梯问题的一个二维变种，这里使用二维dp来求解。

第一次提交，只超过了10%。

class Solution(object): def uniquePaths(self, m, n): """ :type m: int :type n: int :rtype: int """ ans = [[0 for _ in range(m)] for _ in range(n)] ans[0] = [1 for i in range(m)] for i in range(n): ans[i][0] = 1 for i in range(1, n): for j in range(1, m): ans[i][j] = ans[i-1][j] + ans[i][j-1] return ans[n-1][m-1]

第二次，直接生成对应的数组，超过80%。

class Solution(object): def uniquePaths(self, m, n): """ :type m: int :type n: int :rtype: int """ ans = [[0 if i!=0 else 1 for i in range(m)] if j != 0 else [1]*m for j in range(n)] for i in range(1, n): for j in range(1, m): ans[i][j] = ans[i-1][j] + ans[i][j-1] return ans[n-1][m-1]

第三次，优化行的生成方式，超过了95%的提交。

class Solution(object): def uniquePaths(self, m, n): """ :type m: int :type n: int :rtype: int """ ans = [[1]+[0]*(m-1) if j != 0 else [1]*m for j in range(n)] for i in range(1, n): for j in range(1, m): ans[i][j] = ans[i-1][j] + ans[i][j-1] return ans[n-1][m-1]

总结

建立整个数组后再逐个修改最慢在列表推导式中使用if-else次之在行的生成中去掉if-else最快（[0]*m比[0 for _ in range(m)]也快一点）

更多参考列表推导式