/**
 * Created by lvhao on 2017/7/6.
 * 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?
 动态规划，目标是到最终点的路径有几条。每个点的路径都是能到这个点的上一个点的路径之和，由于
 只能向下和向右走，所以动态方程是A[i][j] = A[i][j-1] + A[j][j+1]
 算出每个点的值就行，最后返回A[m-1][n-1]
 */
public class Q62UniquePaths {
    public static void main(String[] args) {
        System.out.println(uniquePaths(3,3));
    }
    public static int uniquePaths(int m, int n) {
        if (m == 1 || n == 1)
            return 1;
        int[][] res = new int[m][n];
        //一开始要先把二维数组上和左边的初始化为1，这是初始条件
        for (int i = 0; i < m; i++) {
            res[i][0] = 1;
        }
        for (int i = 0; i < n; i++) {
            res[0][i] = 1;
        }
        //动态规划
        for (int i = 1; i < m; i++) {
            for (int j = 1;j < n;j++)
            {
                res[i][j] = res[i-1][j] + res[i][j-1];
            }

        }
        return res[m-1][n-1];
    }

}