[Swift Weekly Contest 109]LeetCode935. 骑士拨号器 | Knight Dialer

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
  

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国际象棋中的骑士可以按下图所示进行移动：

 .           

这一次，我们将 “骑士” 放在电话拨号盘的任意数字键（如上图所示）上，接下来，骑士将会跳 N-1 步。每一步必须是从一个数字键跳到另一个数字键。

每当它落在一个键上（包括骑士的初始位置），都会拨出键所对应的数字，总共按下 N 位数字。

你能用这种方式拨出多少个不同的号码？

因为答案可能很大，所以输出答案模 10^9 + 7。

示例 1：

输入：1
输出：10

示例 2：

输入：2
输出：20

示例 3：

输入：3
输出：46

提示：

• 1 <= N <= 5000

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284ms

 1 class Solution {
 2     func knightDialer(_ N: Int) -> Int {
 3         var N = N
 4         var mod:Int64 = 1000000007
 5         N -= 1
 6         var dp:[Int64] = [Int64](repeating: 1,count: 10)
 7         for i in 0..<N
 8         {
 9             var ndp:[Int64] = [Int64](repeating: 0,count: 10)
10             ndp[0] = dp[4] + dp[6]
11             ndp[1] = dp[6] + dp[8]
12             ndp[2] = dp[7] + dp[9]
13             ndp[3] = dp[4] + dp[8]
14             ndp[4] = dp[3] + dp[9] + dp[0]
15             ndp[6] = dp[1] + dp[7] + dp[0]
16             ndp[8] = dp[1] + dp[3]
17             ndp[7] = dp[2] + dp[6]
18             ndp[9] = dp[2] + dp[4]
19             for j in 0..<10
20             {
21                 ndp[j] %= mod
22             }
23             dp = ndp
24         }
25         var ret:Int64 = 0
26         for i in 0..<10
27         {
28             ret += dp[i]
29         }
30         return Int(ret % mod)
31     }
32 }