## 题目描述：

LeetCode 552. Student Attendance Record II

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:

1. 'A' : Absent.
2. 'L' : Late.
3. '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.

## 解题思路：

`利用dp[n][A][L]表示长度为n，包含A个字符'A'，以L个连续的'L'结尾的字符串的个数。`

```dp[n][0][0] = sum(dp[n - 1][0])
dp[n][0][1] = dp[n - 1][0][0]
dp[n][0][2] = dp[n - 1][0][1]
dp[n][1][0] = sum(dp[n - 1][0]) + sum(dp[n - 1][1])
dp[n][1][1] = dp[n - 1][1][0]
dp[n][1][2] = dp[n - 1][1][1]

## Java代码：

``````public class Solution {
private final int MOD = 1000000007;
public long sum(int[] nums) {
long ans = 0;
for (int n : nums) ans += n;
return ans % MOD;
}

public int checkRecord(int n) {
int dp[][] = {{1, 1, 0}, {1, 0, 0}};
for (int i = 2; i <= n; i++) {
int ndp[][] = {{0, 0, 0}, {0, 0, 0}};
ndp[0][0] = (int)sum(dp[0]);
ndp[0][1] = dp[0][0];
ndp[0][2] = dp[0][1];
ndp[1][0] = (int)((sum(dp[0]) + sum(dp[1])) % MOD);
ndp[1][1] = dp[1][0];
ndp[1][2] = dp[1][1];
dp = ndp;
}
return (int)((sum(dp[0]) + sum(dp[1])) % MOD);
}

}
``````

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