[Leetcode] 935 Knight Dialer dialer Knight

Topic:. 935 Knight Dialer Dialer Knight

Problem Description

Chess knight move can press shown below:

This time, we will "knight" in any number key telephone dial pad (shown above), the next hop will knight N-1 step. Each step must be the number keys to jump from one to another number key.
Whenever it falls on a key (including the initial position of the knight), are allocated corresponding to the numeric key, a total of N-bit digital press.
You can use this way how many different numbers dialed?
Because the answer may be large, so the output answer mode 10 ^ 9 + 7 .

• Example 1:

  Input: 1
  Output: 10

• Example 2:

  Input: 2
  Output: 20

• Example 3:

  Input: 3
  Output: 46

• prompt:

  1 <= N <= 5000

Ideas 1

First, we look at the knight to walk on the dialpad

Thought the use of dynamic programming, are each a result of a computation obtained by the above (except for the first round)

Algorithm

• Not by OJ
• The idea is correct, limited data types, obtaining greater value will overflow

class Solution {
    public int knightDialer(int N) {
        if (N == 1) return 10;
        
        // 第二轮 每个键位下一步数
        int[] jumpSizes = {2,2,2,2,3,0,3,2,2,2};
        // 临时存放 运算的值
        int[] tempSizes = new int[10];
        
        // 从底层开始 推理出顶层的答案
        for (int i = 2; i < N; i++) {
            giveNum(tempSizes,0, jumpSizes, 4, 6);
            giveNum(tempSizes,1, jumpSizes, 6, 8);
            giveNum(tempSizes,2, jumpSizes, 7, 9);
            giveNum(tempSizes,3, jumpSizes, 4, 8);
            giveNum(tempSizes,4, jumpSizes, 3, 9, 0);
            giveNum(tempSizes,6, jumpSizes, 1, 7, 0);
            giveNum(tempSizes,7, jumpSizes, 2, 6);
            giveNum(tempSizes,8, jumpSizes, 1, 3);
            giveNum(tempSizes,9, jumpSizes, 2, 4);
            tempSizes[5] = 0;
            
            // 更新当前轮次的结果
            jumpSizes = Arrays.copyOf(tempSizes, 10);
        }
        
        int sum = 0;
        for (int num : jumpSizes) sum += num;
        
        return sum;
    }
    
    void giveNum(int[] dest, int i, int[] src, int j, int k) {
        dest[i] = src[j] + src[k];
    }
    
    void giveNum(int[] dest, int i, int[] src, int j, int k, int p) {
        dest[i] = src[j] + src[k] + src[p];
    }
}

• By OJ
• Process is the same, but changing the type of data , plus modulo

class Solution {
    int mod = 1000000007;
    public int knightDialer(int N) {
        if (N == 1) return 10;
        
        // 将int改为long 数据不会溢出
        long[] jumpSizes = {2,2,2,2,3,0,3,2,2,2};
        long[] tempSizes = new long[10];
        long sum = 0;
        
        for (int i = 2; i < N; i++) {
            giveNum(tempSizes,0, jumpSizes, 4, 6);
            giveNum(tempSizes,1, jumpSizes, 6, 8);
            giveNum(tempSizes,2, jumpSizes, 7, 9);
            giveNum(tempSizes,3, jumpSizes, 4, 8);
            giveNum(tempSizes,4, jumpSizes, 3, 9, 0);
            giveNum(tempSizes,6, jumpSizes, 1, 7, 0);
            giveNum(tempSizes,7, jumpSizes, 2, 6);
            giveNum(tempSizes,8, jumpSizes, 1, 3);
            giveNum(tempSizes,9, jumpSizes, 2, 4);
            tempSizes[5] = 0;
            
            jumpSizes = Arrays.copyOf(tempSizes, 10);
        }
        
        for (long num : jumpSizes) sum += num;
        return (int)(sum%mod);
    }
    
    void giveNum(long[] dest, int i, long[] src, int j, int k) {
        // 每次求和后 都进行取模
        dest[i] = (src[j] + src[k]) % mod;
    }
    
    void giveNum(long[] dest, int i, long[] src, int j, int k, int p) {
        dest[i] = (src[j] + src[k] + src[p]) % mod;
    }
}

Ideas 2

By observing the position of all the buttons and the calculation of the trajectory, summed Law

Algorithm

class Solution {
    public int knightDialer(int N) {
        if (N == 1) return 10;
        
        int mod = 1000000007;
        // a表示四角的和
        // b表示中部左右的和
        // c表示中部上下的和
        // d表示0键位的值
        long a = 4, b = 2, c = 2, d = 1;
        for (int i = 1; i < N; i++) {
            long A = (2*(b+c)) % mod;
            long B = (a + 2*d) % mod;
            c = a;
            d = b;
            a = A;
            b = B;
        }
        
        return (int)((a+b+c+d)%mod);
    }
}