[LeetCode] 925. Long Pressed Name

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.

Constraints:

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

你的朋友正在使用键盘输入他的名字 name。偶尔，在键入字符 c 时，按键可能会被长按，而字符可能被输入 1 次或多次。

你将会检查键盘输入的字符 typed。如果它对应的可能是你的朋友的名字（其中一些字符可能被长按），那么就返回 True。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/long-pressed-name
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路是two pointer，一个i指针走name，一个j指针走typed。当没有遍历完name的时候，如果两边的字符是一样的，则往前走；如果两边字符不一样，则需要判断

j是否在0的位置上，如果是，那说明第一个字母就对不上，直接return false
j如果不在0的位置上，则判断j是否跟j - 1位置上的字母相等，如是，则有可能是重复字母

时间O(n)

空间O(1)

Java实现

 1 class Solution {
 2     public boolean isLongPressedName(String name, String typed) {
 3         int i = 0;
 4         int m = name.length();
 5         int n = typed.length();
 6         for (int j = 0; j < n; j++) {
 7             if (i < m && name.charAt(i) == typed.charAt(j)) {
 8                 i++;
 9             } else if (j == 0 || typed.charAt(j) != typed.charAt(j - 1)) {
10                 return false;
11             }
12         }
13         return i == m;
14     }
15 }

LeetCode 题目总结