2018/11/11

935. Knight Dialer

問題描述：

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-1hops. 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

How many distinct numbers can you dial in this manner?

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

樣例：

Example 1:

Input: 1
Output: 10

Example 2:

Input: 2
Output: 20

Example 3:

Input: 3
Output: 46

Note:

• 1 <= N <= 5000

問題分析：

本題難度為Medium！屬於動態規劃問題，已給出的函式定義為

class
 
 Solution:
    def knightDialer(self, N):
        """
        :type N: int
        :rtype: int
        """

本題採取動態規劃演算法，每一步的電話數字與上一步的電話數字相關。nextNumbers本質上為一張查詢表，nextNumber[i]儲存了數字i下一步可以到達的數字；res[i]表示最後一步停留在數字i上的組合數（情況數），則sum(res)為最後輸出，每一步都更新新的結果res=newRes；newRes是在上一步res的基礎上當前這一步的結果。

程式碼實現：

#coding=utf-8
 

class Solution:
    def knightDialer(self, N):
        """
        :type N: int
        :rtype: int
        """
        #當前數字可以跳到的下一個數字，例：nextNumbers[0]=[4,6]表示0的可以跳到數字4或6
        nextNumbers=[[4,6],[6,8],[7,9],[4,8],[3,9,0],[],[1,7,0],[2,6],[1,3],[2,4]] 
        res = [1 for t in range(10)] #某一步數將走上其索引所在數字的所有情況的總數，初始化為第一步的結果
        
        for i in range(1,N): #對於第i步
            newRes = [0 for t in range(10)] #用來更新其總數情況。
            for number in range(10):
                for j in nextNumbers[number]:
                    newRes[j] += res[number]
            res = newRes
        return sum(res) % (10**9+7) #對輸出取模