# 925.Long Pressed Name

**925.Long Pressed Name**

难度:Easy

> Your friend is typing his name into a keyboard. Sometimes, when typing a character c, the key might get long pressed, and the character will be typed 1 or more times.

You examine the typed characters of the keyboard. Return True if it is possible that it was your friends name, with some characters (possibly none) being long pressed.

```
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:

name.length <= 1000
typed.length <= 1000
The characters of name and typed are lowercase letters.
```

首先可以確定如果輸入的字串長度小於name的話，直接返回false。另外從name的起點來時判斷，對輸入字串遍歷一遍，如果前一個和後一個相等，則看name的前後是否相等，如果相等，則計數器start後移，如果不相等則保持不變。如果輸入字串不相等，也直接加1.碰到name和輸入字串相應位置不等的情況，直接返回false。最後看start是否走到了name的末尾，如果沒有就返回false，否則返回true。

```
class Solution {
public:
    bool isLongPressedName(string name, string typed) {
        if(typed.length()<name.length()) return false;
        int start=0;
        for(int i=0;i<typed.size();i++)
        {
            if(typed[i]!=name[start]) return false;
            else if(i<typed.size()-1 && typed[i] == typed[i+1]){
                if(start<name.size()-1 && name[start] == name[start+1])
                start++;
            }
            else 
                start++;
        }
        return start==name.size();
    }
};
```

> 执行用时 :0 ms, 在所有 C++ 提交中击败了100.00%的用户\
> 内存消耗 :8.5 MB, 在所有 C++ 提交中击败了77.52%的用户