552. Student Attendance Record II

最新推荐文章于 2021-08-18 01:07:42 发布

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最新推荐文章于 2021-08-18 01:07:42 发布

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本文介绍了一种算法，用于计算长度为n的所有可能的学生考勤记录数，这些记录被认为是值得奖励的，即最多只有一天缺席且没有连续三天迟到的情况。通过动态规划（DP）的方法进行求解，并给出具体实现代码。

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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.

DP求解，详细的分析见链接详细说明，程序如下所示：

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