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?
Note: m and n will be at most 100.
Example 1:
Input: m = 3, n = 2Output: 3Explanation:From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:1. Right -> Right -> Down2. Right -> Down -> Right3. Down -> Right -> Right
Example 2:
Input: m = 7, n = 3Output: 28 我们需要用动态规划Dynamic Programming来解，我们可以维护一个二维数组dp，其中dp[i][j]表示到当前位置不同的走法的个数，然后可以得到递推式为: dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
class Solution { public int uniquePaths(int m, int n) { if(m < 0 || n < 0){ return -1; } int[][] dp = new int[m][n]; for(int i = 0; i < m; i++){ for(int j = 0; j < n; j++){ if(i == 0){ dp[i][j] = 1; } else if(j == 0){ dp[i][j] = 1; } else{ dp[i][j] = dp[i-1][j] + dp[i][j-1]; } } } return dp[m-1][n-1]; }}