925. Long Pressed Name

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：

 class Solution {
 public:
     bool isLongPressedName(string name, string typed) {
         if (name.length() > typed.length()) return false;

         char pre = ' ';
         int indexOfName = ;

         for (int i = ; i < typed.length(); ++i)
         {
             if (name[indexOfName] != typed[i] && pre == ' ')
                 return false;

             if (name[indexOfName] == typed[i])
             {
                 pre = name[indexOfName];
                 indexOfName++;
             }
             else
             {
                 if (pre == typed[i])
                     continue;
                 else
                     return false;
             }
         }

         if (indexOfName != name.length()) return false;

         return true;
     }
 };