Level：
Medium
题目描述：
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 7 x 3 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
Example 1:
Input: m = 3, n = 2
Output: 3
Explanation:
From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right
Example 2:
Input: m = 7, n = 3
Output: 28
思路分析：
题目要求求出一个机器人从矩阵的左上角走到矩阵的右下角，一共有多少种走法。可以用动态规划的思想来解决这道题，我们用dp[ i ] [ j ]来表示走到第i行和第j列，共有多少种走法。由题意知，机器人只能向右和向下走，那么状态转移方程是 dp[ i ] [ j ]=dp [i-1] [ j ]+dp[ i ] [ j-1]。注意到矩阵第一行或者第一列某个位置，路径只有一条。（因为起点是左上角，并且只能向右向下移动）
代码：
public class Solution{
public int uniquePaths(int m,int n){
int [][]dp=new int [m][n];
for(int i=0;i<m;i++){
dp[i][0]=1;
}
for(int j=0;j<n;j++){
dp[0][j]=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];
}
}