Content

There is a robot on an m x n grid. The robot is initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]). The robot can only move either down or right at any point in time.

Given the two integers m and n, return the number of possible unique paths that the robot can take to reach the bottom-right corner.

The test cases are generated so that the answer will be less than or equal to 2 * 10⁹.

Example 1:

Input: m = 3, n = 7
Output: 28

Example 2:

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 -> Down -> Down
2. Down -> Down -> Right
3. Down -> Right -> Down

Constraints:

1 <= m, n <= 100

Solution

1. 动态规划

Java

class Solution {
    public int uniquePaths(int m, int n) {
        // 1 <= m, n <= 100
        // dp[i][j]表示 i * j 的 grid 的可能路径数
        int[][] dp = new int[m + 1][n + 1];
        Arrays.fill(dp[1], 1);
        for (int i = 1; i <= m; i++) {
            dp[i][1] = 1;
        }
        for (int i = 2; i <= m; i++) {
            for (int j = 2; j <= n; j++) {
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
        return dp[m][n];
    }
}