521. Longest Uncommon Subsequence I

Given a group of two strings, you need to find the longest uncommon subsequence of this group of two strings. The longest uncommon subsequence is defined as the longest subsequence of one of these strings and this subsequence should not be any subsequence of the other strings.

A subsequence is a sequence that can be derived from one sequence by deleting some characters without changing the order of the remaining elements. Trivially, any string is a subsequence of itself and an empty string is a subsequence of any string.

The input will be two strings, and the output needs to be the length of the longest uncommon subsequence. If the longest uncommon subsequence doesn’t exist, return -1.

Example 1:

Input: "aba", "cdc"
Output: 3
Explanation: The longest uncommon subsequence is "aba" (or "cdc"), 
because "aba" is a subsequence of "aba", 
but not a subsequence of any other strings in the group of two strings.

Note:

1. Both strings’ lengths will not exceed 100.
2. Only letters from a ~ z will appear in input strings.

Solution in C++：

關鍵點：

• 分類討論

思路：

1. 哇，開始還以為很難，最開始寫的程式碼只是處理一下我認為的特殊情況，然後在想其他情況的case的時候，我想了幾個，比如acb與ab，我以為會輸出2，但是結果不是，我就想到了因為前面不相等直接輸出其中的長度最大值了，然後再就想不出來能跨過這個的例子了，打算submit一下，看有沒有錯誤提示，結果AC了，滿臉震驚，然後去看題解居然也是這樣做的，看了分析才認識到真正的原理。真是分析大法好啊。
   解析

int findLUSlength(string a, string b) {
        if (a != b)
            return max(a.size(), b.size());
        else
            return -1;
    }

551. Student Attendance Record I

You are given a string representing an attendance record for a student. The record only contains the following three characters:

1. ’A’ : Absent.
2. ’L’ : Late.
3. ’P’ : Present.
   A student could be rewarded if his attendance record doesn’t contain more than one ‘A’ (absent) or more than two continuous ‘L’ (late).

You need to return whether the student could be rewarded according to his attendance record.

Example 1:

Input: "PPALLP"
Output: True

Example 2:

Input: "PPALLL"
Output: False

Solution in C++：

關鍵點：

• 連續的L

思路：

• 開始沒看到連續，然後老是錯，還以為自己邏輯看錯了，後來發現是題意理解錯了（不過也是通過例子才知道的），接下來就是連續L計數的問題，開始把不是連續L的值設定為0出現了邏輯問題，後來就把不是連續的L設定成1就解決了。大框架就是遍歷就好。然後可能這裡還有就是a和l的值判斷的時機應該放在改變值之後，以免遇到字串結尾才出現False的情況。
• 看到discuss裡面判斷連續L的邏輯是碰到P之後將l的值歸零，其他情況正常+1，感覺這個邏輯也蠻不錯的，點個贊。

bool checkRecord(string s) {
        int a = 0;
        int l = 0;
        
        for(int i = 0; i < s.size(); ++i){
            if (s[i] == 'A')
                ++a;
            else if (s[i] == 'L'){
                if (i != 0 && s[i-1] == 'L')
                    ++l;
                else
                    l = 1;
            }
            if (a > 1)
                return false;
            if (l > 2)
                return false;
        }
        
        return true;
    }

小結

今天實驗室出去聚會了，回來比較晚了，明天還有課還要去實驗室打卡，今天就刷不到那麼長時間了，但是不過卻有不少收穫，思維上的鍛鍊得到了，很享受。

知識點

• 分類討論