LeetCode

题目：

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 3 x 7 grid. How many possible unique paths are there?

思路：

动态规划，到达某一点的路线数是其上和左的节点的路线数之和。我们也可以用C(M+N-2, M-1)来计算，即C(8, 2) = 28.

package dp;

public class UniquePaths {

    public int uniquePaths(int m, int n) {
        int[][] dp = new int[m][n];
        dp[0][0] = 1;
        for (int i = 1; i < m; ++i) dp[i][0] = 1;
        for (int i = 1; i < n; ++i) dp[0][i] = 1;
        for (int i = 1; i < m; ++i) {
            for (int j = 1; j < n; ++j) {
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
        return dp[m - 1][n - 1];
    }
    
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        UniquePaths u = new UniquePaths();
        System.out.println(u.uniquePaths(3, 7));
    }

}