(HW)uniquePath(Java)

public class test
{
    public static void main(String[] args)
    {
        Scanner input = new Scanner(System.in);
        int m = input.nextInt();
        int n = input.nextInt();
        //iterative
        System.out.println(uniquePath(m, n));
        //recursive
        int[][] dp = new int[m][n];
        System.out.println(uniquePath(m - 1, n - 1, dp));
    }    
    
    //iterative
    public static int uniquePath(int m, int n)  //时间复杂度: O(n^m)
    {
        int[][] dp = new int[m][n];
        
        //第一行均赋为1
        for(int j = 0; j < dp[0].length; j++)
            dp[0][j] = 1;
        
        //第一列均赋为1
        for(int i = 0; i < dp.length; i++)
            dp[i][0] = 1;
        
        for(int i = 1; i < dp.length; i++)
            for(int j = 1; j < dp[0].length; j++)
                dp[i][j] = dp[i][j - 1] + dp[i - 1][j];
        
        //返回右下角的值
        return dp[dp.length - 1][dp[0].length - 1];
    }
    
    //recursive
    public static int uniquePath(int m, int n, int[][] dp)
    {
        //递归终止条件
        if(m == 0 || n == 0)
            return 1;
        
        //避免重复计算
        if(dp[m - 1][n] == 0)
            dp[m - 1][n] = uniquePath(m - 1, n, dp);
        if(dp[m][n - 1] == 0)
            dp[m][n - 1] = uniquePath(m, n - 1, dp);
        dp[m][n] = dp[m - 1][n] + dp[m][n - 1];
        return dp[m][n];
    }
}