552. Student Attendance Record II

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探讨如何通过动态规划算法计算长度为n的所有可能的考勤记录数量，这些记录被视为奖励性的，即最多只能有一次缺席(A)或连续两次迟到(L)。

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Given a positive integer n, return the number of all possible attendance records with length n, which will be regarded as rewardable. The answer may be very large, return it after mod 109 + 7.

A student attendance record is a string that only contains the following three characters:

'A' : Absent.
'L' : Late.
'P' : Present.

A record is regarded as rewardable if it doesn't contain more than one 'A' (absent) or more than two continuous 'L' (late).

Example 1:

Input: n = 2
Output: 8 
Explanation:
There are 8 records with length 2 will be regarded as rewardable:
"PP" , "AP", "PA", "LP", "PL", "AL", "LA", "LL"
Only "AA" won't be regarded as rewardable owing to more than one absent times.

Note: The value of n won't exceed 100,000.

利用动态规划解题。初始化数组P,L,A,PLP,PLL来存储之前的状态，动态规划的原则如下：

PLP[i] = PLL[i - 1] + PLP[i - 1];
PLL[i] = PLP[i - 2] + PLP[i - 1];
P[i] = P[i - 1] + L[i - 1] + A[i - 1];
A[i] = PLP[i - 1] + PLL[i - 1];
L[i] = P[i - 1] + P[i - 2] + A[i - 1] + A[i - 2];

根据以上规则写代码，代码如下：

public class Solution {
    public int checkRecord(int n) {
        if (n == 1) {
            return 3;
        }
        if (n == 2) {
            return 8;
        }
        long[] PLP = new long[n + 1];
        long[] PLL = new long[n + 1];
        long[] P = new long[n + 1];
        long[] L = new long[n + 1];
        long[] A = new long[n + 1];
        PLP[0] = 0; PLP[1] = 1; PLP[2] = 2;
        PLL[0] = 0; PLL[1] = 1; PLL[2] = 2;
        P[0] = 0; P[1] = 1; P[2] = 3;
        L[0] = 0; L[1] = 1; L[2] = 3;
        A[0] = 0; A[1] = 1; A[2] = 2;
        long sumA = 2;
        for (int i = 3; i <= n; i ++) {
            PLP[i] = PLL[i - 1] + PLP[i - 1];
            PLL[i] = PLP[i - 2] + PLP[i - 1];
            P[i] = P[i - 1] + L[i - 1] + A[i - 1];
            A[i] = PLP[i - 1] + PLL[i - 1];
            L[i] = P[i - 1] + P[i - 2] + A[i - 1] + A[i - 2];
            PLP[i] %= 1000000007;
            PLL[i] %= 1000000007;
            P[i] %= 1000000007;
            A[i] %= 1000000007;
            L[i] %= 1000000007;
        }
        return (int)((P[n] + L[n] + A[n]) % 1000000007);
    }
}