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题目：
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.
代码：oj测试通过 Runtime: 44 ms
1 class Solution: 2 # @return an integer 3 def uniquePaths(self, m, n): 4 # none case 5 if m < 1 or n < 1: 6 return 0 7 # special case 8 if m==1 or n==1 : 9 return 1 10 11 # deep first matrix 12 dp = [[0 for col in range(n)] for row in range(m)] 13 # the elements in frist row have only one avaialbe pre-node 14 for i in range(n): 15 dp[0][i]=1 16 # the elements in first column have only one avaialble pre-node 17 for i in range(m): 18 dp[i][0]=1 19 # iterator other elements in the 2D-matrix 20 for row in range(1,m): 21 for col in range(1,n): 22 dp[row][col]=dp[row-1][col]+dp[row][col-1] 23 24 return dp[m-1][n-1]
思路：
动态规划经典题目，用迭代的方法解决。
1. 先处理none case和special case
2. 2D-matrix的第一行和第一列上的元素 只能从上面的元素或左边的元素达到，因此可以直接获得其值
3. 遍历其余的位置：每一个position只能由其左边或者上边的元素达到，这样可得迭代公式 dp[row][col]=dp[row-1][col]+dp[row][col-1]
4. 遍历完成后 dp矩阵存放了从其实位置到当前位置的所有可能走法，因此返回dp[m-1][n-1]就是需要的值
leetcode 【 Unique Paths 】python 实现
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原文地址：http://www.cnblogs.com/xbf9xbf/p/4250359.html
