题目链接：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，则typed 的 length(长度)满足条件：typed.length() >= name.length();
• i, j分别是指向name 和 typed的下标，i, j下标初始值都为0；

1. 当 name[i] == typed[j] 时，i, j 向后移动一个单位；
2. 当 name[i] != typed[j] 时，判断 typed[j] 是否等于name[i-1] (name[i-1] == typed[j-1])；

• • 若 typed[j] != name[j-1] 则 typed 不满足，返回false。
  • 若 typed[j] == name[i-1]，则 ++j，直至 typed[j] != name[i-1]，执行步骤2。

 注意：根据上面的分析，需要用一个字符变量来存储name[i]的值，该字符变量初始化为空字符。

编码如下：

 1 class Solution {
 2 public:
 3     bool isLongPressedName(string name, string typed) {
 4         if (name.length() > typed.length()) return false;
 5         
 6         char pre = ' ';
 7         int indexOfName = 0;
 8         
 9         for (int i = 0; i < typed.length(); ++i)
10         {
11             if (name[indexOfName] != typed[i] && pre == ' ')
12                 return false;
13             
14             if (name[indexOfName] == typed[i])
15             {
16                 pre = name[indexOfName];
17                 indexOfName++;
18             }
19             else
20             {
21                 if (pre == typed[i])
22                     continue;
23                 else
24                     return false;
25             }
26         }
27         
28         if (indexOfName != name.length()) return false;
29         
30         return true;  
31     }
32 };