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
难度：medium
题目：
在m * n 的格子左上角放置一个机器人，机器人在任何时候只能向右或向下移动，机器人尝试移动到格子的最右下角。有多少种可能的走法？思路：动态规划，grid[m][n] = grid[m - 1][n] + grid[m][n - 1]
Runtime: 0 ms, faster than 100.00% of Java online submissions for Unique Paths.
Memory Usage: 23.6 MB, less than 41.57% of Java online submissions for Unique Paths.class Solution { public int uniquePaths(int m, int n) { int[][] grid = new int[m][n]; for (int i = 0; i < n; i++) { grid[0][i] = 1; } for (int i = 0; i < m; i++) { grid[i][0] = 1; } for (int i = 1; i < m; i++) { for (int j = 1; j < n; j++) { grid[i][j] = grid[i - 1][j] + grid[i][j - 1]; } } return grid[m - 1][n - 1]; }}