Lintcode：不同的路径

问题：

有一个机器人的位于一个 m × n 个网格左上角。机器人每一时刻只能向下或者向右移动一步。机器人试图达到网格的右下角。

问有多少条不同的路径？

样例：

Example 1:

Input: n = 1, m = 3
Output: 1	
Explanation: Only one path to target position.

Example 2:

Input:  n = 3, m = 3
Output: 6	
Explanation:
	D : Down
	R : Right
	1) DDRR
	2) DRDR
	3) DRRD
	4) RRDD
	5) RDRD
	6) RDDR

注意事项

n和m均不超过100

python：

class Solution:
    """
    @param m: positive integer (1 <= m <= 100)
    @param n: positive integer (1 <= n <= 100)
    @return: An integer
    """
    def uniquePaths(self, m, n):
        # write your code here
        if m == 1 or n == 1:
            return 1
        nums = [[0 for i in range(n)] for j in range(m)]
        for i in range(m):
            nums[i][0] = 1
        for j in range(n):
            nums[0][j] = 1
        for i in range(1,m):
            for j in range(1,n):
                nums[i][j] = nums[i-1][j] + nums[i][j-1]
        return nums[m-1][n-1]

C++：

class Solution {
public:
    /**
     * @param m: positive integer (1 <= m <= 100)
     * @param n: positive integer (1 <= n <= 100)
     * @return: An integer
     */
    int uniquePaths(int m, int n) {
        // write your code here
        if(m == 1 || n == 1)
        {
            return 1;
        }
        vector<vector<int>> nums(m,vector<int>(n));
        for(int i = 1; i < m; i++)
        {
            nums[i][0] = 1;
        }
        for(int j = 1; j < n; j++)
        {
            nums[0][j] = 1;
        }
        for(int i = 1; i < m; i++)
        {
            for(int j = 1; j < n; j++)
            {
                nums[i][j] = nums[i-1][j] + nums[i][j-1];
            }
        }
        return nums[m-1][n-1];
    }
};