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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
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，gridm = gridm – 1 + gridm
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];
}
}