LeetCode 935. Knight Dialer

原题链接在这里：https://leetcode.com/problems/knight-dialer/

题目：

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

题解：

The question asks distinct numbers could dial.

It is actually the sum of ways jump ending at each cell.

Cell 1 could jump to cell 6 and 8. Thus accumlate the current ways count to next ways at 6 and 8.

Eventually, get all the sum.

Time Complexity: O(N).

Space: O(1).

AC Java: 

class Solution {
    public int knightDialer(int N) {
        if(N == 0){
            return 0;
        }
        
        if(N == 1){
            return 10;
        }
        
        int M = 1000000007;
        long [] cur = new long[10];
        Arrays.fill(cur, 1);
        for(int k = 2; k<=N; k++){
            long [] next = new long[10];

            next[1] = (cur[6]+cur[8])%M;
            next[2] = (cur[7]+cur[9])%M;
            next[3] = (cur[4]+cur[8])%M;
            next[4] = (cur[3]+cur[9]+cur[0])%M;
            next[5] = 0;
            next[6] = (cur[1]+cur[7]+cur[0])%M;
            next[7] = (cur[2]+cur[6])%M;
            next[8] = (cur[1]+cur[3])%M;
            next[9] = (cur[2]+cur[4])%M;
            next[0] = (cur[4]+cur[6])%M;
            
            cur = next;
        }
        
        long res = 0;
        for(int i = 0; i<10; i++){
            res = (res + cur[i]) % M;
        }
        return (int)res;
    }
}

类似Number of Ways to Stay in the Same Place After Some Steps.