1 public class test
2 {
3 public static void main(String[] args)
4 {
5 Scanner input = new Scanner(System.in);
6 int m = input.nextInt();
7 int n = input.nextInt();
8 //iterative
9 System.out.println(uniquePath(m, n));
10 //recursive
11 int[][] dp = new int[m][n];
12 System.out.println(uniquePath(m - 1, n - 1, dp));
13 }
14
15 //iterative
16 public static int uniquePath(int m, int n) //时间复杂度: O(n^m)
17 {
18 int[][] dp = new int[m][n];
19
20 //第一行均赋为1
21 for(int j = 0; j < dp[0].length; j++)
22 dp[0][j] = 1;
23
24 //第一列均赋为1
25 for(int i = 0; i < dp.length; i++)
26 dp[i][0] = 1;
27
28 for(int i = 1; i < dp.length; i++)
29 for(int j = 1; j < dp[0].length; j++)
30 dp[i][j] = dp[i][j - 1] + dp[i - 1][j];
31
32 //返回右下角的值
33 return dp[dp.length - 1][dp[0].length - 1];
34 }
35
36 //recursive
37 public static int uniquePath(int m, int n, int[][] dp)
38 {
39 //递归终止条件
40 if(m == 0 || n == 0)
41 return 1;
42
43 //避免重复计算
44 if(dp[m - 1][n] == 0)
45 dp[m - 1][n] = uniquePath(m - 1, n, dp);
46 if(dp[m][n - 1] == 0)
47 dp[m][n - 1] = uniquePath(m, n - 1, dp);
48 dp[m][n] = dp[m - 1][n] + dp[m][n - 1];
49 return dp[m][n];
50 }
51 }
