62. Unique Paths

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?

Above is a 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

如题，首先最先想到的是数学方法，m行n列，一共需要走m+n-2步，其中m-1步是横向移动，所以只要求C(m+n-2,m-1)，用递归的方法

class Solution(object):    def jiecheng(self,n):        if n==1 or n==0:            return 1        else:            return self.jiecheng(n-1)*n    def uniquePaths(self, m, n):        """        :type m: int        :type n: int        :rtype: int        """        return self.jiecheng(m+n-2)/(self.jiecheng(m-1)*self.jiecheng(n-1))

考虑动态规划，从0,0走到位置i，j有两种方式，一种是到达了i-1，j，另一种是到达了i，j-1.所需步数p[i][j]等于从0,0走到i-1，j的步数加上从0,0走到i，j-1的步数之和即p[i][j]=p[i-1][j]+p[i][j-1]

代码如下：

class Solution:    # @return an integer    def uniquePaths(self, m, n):        aux = [[1 for x in range(n)] for x in range(m)]        for i in range(1, m):            for j in range(1, n):                aux[i][j] = aux[i][j-1]+aux[i-1][j]        return aux[-1][-1]