62. Unique Paths

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?

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

Note: m and n will be at most 100.

一共有两种方法，第一种很容易想到的是递归，即坐标（3,7）处的值是（3,6）处的值与（2,7）处的值之和，并且对于第一行以及第一列的值都为1，写出递归算法的解

class Solution {public:    int uniquePaths(int m, int n) {        if(1 == m)            return 1;        if(1 == n)            return 1;        if(m>1 && n>1)        {            return uniquePaths(m-1, n) + uniquePaths(m, n-1);        }    }};

可惜提交后显示超时。

第二种方法和递归思路一致，只不过是进行两层for循环来把每个坐标位置的值都计算出来，最后返回最右下脚坐标的值。

class Solution {public:    int uniquePaths(int m, int n) {        vector<vector<int> > path(m, vector<int>(n, 1));        for(int i = 1; i < m; i ++)        {            for(int j = 1; j < n; j ++)            {                path[i][j] = path[i-1][j] + path[i][j-1];            }        }        return path[m-1][n-1];    }};