題目如下：

A chess knight can move as indicated in the chess diagram below:

 .           

 

This time, we place our chess knight on any numbered key of a phone pad (indicated above), and the knight makes N-1 hops.  Each hop must be from one key to another numbered key.

Each time it lands on a key (including the initial placement of the knight), it presses the number of that key, pressing N digits total.

How many distinct numbers can you dial in this manner?

Since the answer may be large, output the answer modulo 10^9 + 7.

Example 1:

Input: 1
Output: 10

Example 2:

Input: 2
Output: 20

Example 3:

Input: 3
Output: 46

Note:

• 1 <= N <= 5000

解題思路：很明顯的動態規劃的場景。首先我們可以維護如下的一個對映字典key為到達的數字，value為可以由哪些數字經過一次跳躍到達key的數字。接下來假設dp[i][j] 為經過跳躍i次並且最後一次跳躍的終點是j，那麼有dp[i][j] = dp[i-1][dic[j][0]] + dp[i-1][dic[j][1]] + ... dp[i-1][dic[j][n]]。最終的結果就是dp[N][0] + dp[N][1] + ... dp[N][9]。後來經過測試發現，這種解法會超時，因為題目約定了N最大是5000，因此可以事先計算出1~5000的所有結果快取起來。

　　　　 dic[1] = [6,8]
        dic[2] = [7,9]
        dic[3] = [4,8]
        dic[4] = [3,9,0]
        dic[5] = []
        dic[6] = [1,7,0]
        dic[7] = [2,6]
        dic[8] = [1,3]
        dic[9] = [2,4]
        dic[0] = [4,6]

程式碼如下：

class Solution(object):
    res = []
    def knightDialer(self, N):
        """
        :type N: int
        :rtype: int
        """
        if len(self.res) != 0:
            return self.res[N-1]
        dic = {}
        dic[1] = [6,8]
        dic[2] = [7,9]
        dic[3] = [4,8]
        dic[4] = [3,9,0]
        dic[5] = []
        dic[6] = [1,7,0]
        dic[7] = [2,6]
        dic[8] = [1,3]
        dic[9] = [2,4]
        dic[0] = [4,6]
        dp = []
        for i in range(5001):
            if i == 0:
                tl = [1] * 10
            else:
                tl = [0] * 10
            dp.append(tl)
        for i in range(5001):
            for j in range(10):
                for k in dic[j]:
                    dp[i][j] += dp[i-1][k]
        for i in range(5001):
            self.res.append(sum(dp[i]) % (pow(10,9) + 7))
        return self.res[N-1]