https://leetcode-cn.com/problems/unique-paths/

法一：排列组合

public int uniquePaths(int m, int n) {
        long ans = 1;
        for (int x = n, y = 1; y < m && x <= m + n - 2; ++x, ++y) {
            ans = ans * x / y;
        }
        return (int) ans;  
}

排列组合问题。。。。
从（0， 0）往下最多能走5步，往右最多能走4步，

法二：动态规划

public int uniquePaths(int m, int n) {
    return process(0, 0, m, n);
}
private int process(int row, int col, int m, int n) {
    if (row == m - 1 && col == n - 1) {
        return 1;
    }else if (row == m - 1) {
        return 1;
    }else if (col == n - 1) {
        return 1;
    }else if (row > m - 1 || col > n - 1) {
        return 0;
    }
    return process(row, col + 1, m, n) + process(row + 1, col, m, n);
}

 public int uniquePaths(int m, int n) {
     int[][] dp = new int[m][n];
     for (int i = n - 1; i >= 0; i--) {
         dp[m - 1][i] = 1;
     }
     for (int i = 0; i < m; i++) {
         dp[i][n - 1] = 1;
     }
     for (int i = m - 2; i >= 0; i--) {
         for (int j = n - 2; j >= 0; j--) {
             dp[i][j] = dp[i][j + 1] + dp[i + 1][j];
         }
     }
     return dp[0][0];
 }