内容简介: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 diag
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,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];
}
}
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