LeetCode-Long Pressed Name

最新推荐文章于 2020-10-21 11:41:01 发布

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本文介绍了一种算法，用于判断在用户可能长按键盘字符的情况下，输入的字符串是否能匹配预设的名字字符串。通过遍历名字字符串并与输入字符串进行对比，确保即使有字符被长按，也能正确识别。

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Description：
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.

题意：给定一个字符串name，及一个用户敲击的字符串typed；由于用户敲击键盘中的一个字符时可能时间较长，导致一个字符出现多次；现在考虑敲击时间过长的这种情况下，判断字符串typed是否与字符串name相符；

解法：遍历字符串name，假设遍历name的当前位置为i，我们只需要从字符串typed的当前位置出发，找到与字符name.charAt(i)相等的那个字符（如果存在的话）；

Java

class Solution {
    public boolean isLongPressedName(String name, String typed) {
        int index = 0;
        for (int i = 0; i < name.length(); i++) {
            while (index < typed.length() && name.charAt(i) != typed.charAt(index)) index++;
            if (index >= typed.length()) return false;
            index++;
        }
        return true;
    }
}