題目連結：https://leetcode.com/problems/long-pressed-name/description/

Example 1:

Input: name = "alex", typed = "aaleex"
Output: true
Explanation: 'a' and 'e' in 'alex' were long pressed.

Example 2:

Input: name = "saeed", typed = "ssaaedd"
Output: false
Explanation: 'e' must have been pressed twice, but it wasn't in the typed output.
 

Example 3:

Input: name = "leelee", typed = "lleeelee"
Output: true

Example 4:

Input: name = "laiden", typed = "laiden"
Output: true
Explanation: It's not necessary to long press any character.

Note:

1. name.length <= 1000
2. typed.length <= 1000
3. The characters of name
    and typed are lowercase letters.

思路：

• 若typed 符合要求，則typed 的 length(長度)滿足條件：typed.length() >= name.length();
• i, j分別是指向name 和 typed的下標，i, j下標初始值都為0；

1. 當 name[i] == typed[j] 時，i, j 向後移動一個單位；
2. 當 name[i] != typed[j] 時，判斷 typed[j] 是否等於name[i-1] (name[i-1] == typed[j-1])；
3. 若 typed[j] != name[j-1] 則 typed 不滿足，返回false。若 typed[j] == name[i-1]，則 ++j，直至 typed[j] != name[i-1]，執行步驟2。

 注意：根據上面的分析，需要用一個字元變數來儲存name[i]的值，該字元變數初始化為空字元。

編碼如下：

 1 class Solution {
 2 public:
 3     bool isLongPressedName(string name, string typed) {
 4         // name的長度 大於 typed的長度時不符合
 5         if (name.length() > typed.length()) return false;
 6         
 7         char pre = ' ';                  // 儲存name[i]的值，初始化為空字元
 8         int indexOfName = 0;      // 指示name的下標   
 9         
10         for (int i = 0; i < typed.length(); ++i)
11         {
12             if (name[indexOfName] != typed[i] && pre == ' ')
13                 return false;
14             
15             if (name[indexOfName] == typed[i])
16             {
17                 pre = name[indexOfName];
18                 indexOfName++;
19             }
20             else
21             {
22                 if (pre == typed[i])
23                     continue;
24                 else
25                     return false;
26             }
27         }
28         
29         if (indexOfName != name.length()) return false;
30         
31         return true;
32         
33     }
34         
35 };