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.

思路：

方法1：思路类似于
http://blog.csdn.net/lanxu_yy/article/details/17501141 方法2：简单思考，每个格子可能的走法就是从左侧的格子或者上面的格子过来。

代码：

方法1：

class Solution {
public:
    int uniquePaths(int m, int n) {
        int ** dp=new int*[m];
        for(int i=0;i<m;i++)
        {
            dp[i] = new int[n];
        }
        
        int step = m+n-1;
        int cur=1;
        dp[0][0]=1;
        while(cur<step)
        {
            for(int i=0;i<=cur;i++)
            {
                if(i>=m)
                {
                    continue;
                }
                int j=cur-i;
                if(j>=n)
                {
                    continue;
                }
                dp[i][j]=0;
                if(i>0)
                {
                    dp[i][j]+=dp[i-1][j];
                }
                if(j>0)
                {
                    dp[i][j]+=dp[i][j-1];
                }
            }
            cur++;
        };
        return dp[m-1][n-1];
    }
};

方法2：

class Solution {
public:
    int uniquePaths(int m, int n) {
        if(m == 0 || n == 0){
            return 0;
        }
        
        int** dp = new int*[m];
        for(int i = 0; i < m; i++){
            dp[i] = new int[n];
            for(int j = 0; j < n; j++){
                dp[i][j] = 0;
            }
        }
        
        
        for(int i = 0; i < m; i++){
            dp[i][0] = 1;
        }
        
        for(int i = 0; i < n; i++){
            dp[0][i] = 1;
        }
        
        for(int i = 1; i < m; i++){
            for(int j = 1; j < n; j++){
                dp[i][j] = dp[i-1][j] + dp[i][j-1];
            }
        }
        
        return dp[m-1][n-1];
    }
};