925. Long Pressed Name

题目链接：https://leetcode.com/problems/long-pressed-name/description/

Example 1:

1Input: name = "alex", typed = "aaleex"
2Output: true
3Explanation: 'a' and 'e' in 'alex' were long pressed.

Example 2:

1Input: name = "saeed", typed = "ssaaedd"
2Output: false
3Explanation: 'e' must have been pressed twice, but it wasn't in the typed output.

Example 3:

1Input: name = "leelee", typed = "lleeelee"
2Output: true

Example 4:

1Input: name = "laiden", typed = "laiden"
2Output: true
3Explanation: 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 1 class Solution {
2 2 public:
3 3     bool isLongPressedName(string name, string typed) {
4 4         if (name.length() > typed.length()) return false;
5 5         
6 6         char pre = ' ';
7 7         int indexOfName = 0;
8 8         
9 9         for (int i = 0; i < typed.length(); ++i)
1010         {
1111             if (name[indexOfName] != typed[i] && pre == ' ')
1212                 return false;
1313             
1414             if (name[indexOfName] == typed[i])
1515             {
1616                 pre = name[indexOfName];
1717                 indexOfName++;
1818             }
1919             else
2020             {
2121                 if (pre == typed[i])
2222                     continue;
2323                 else
2424                     return false;
2525             }
2626         }
2727         
2828         if (indexOfName != name.length()) return false;
2929         
3030         return true;  
3131     }
3232 };