925. Long Pressed Name

最新推荐文章于 2021-05-26 12:17:42 发布

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最新推荐文章于 2021-05-26 12:17:42 发布

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分类专栏： LeedCode365 文章标签： leetcode

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本文介绍了一种算法，用于判断给定的两个字符串中，第二个字符串是否可以通过第一个字符串的字符长按输入得到。通过使用队列数据结构，该算法能够有效识别字符的重复输入，适用于多种应用场景。

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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.

    public boolean isLongPressedName(String name, String typed) {
        Queue<Character> qName = new LinkedList<>();
        Queue<Character> qTyped = new LinkedList<>();
        for (char ch : name.toCharArray()) qName.offer(ch);
        for (char ch : typed.toCharArray()) qTyped.offer(ch);
        while (true) {
            if (qName.peek() == qTyped.peek()) {
                qName.poll();
                qTyped.poll();
                if (qName.size() == 0) return true;
            } else {
                qTyped.poll();
                if (qTyped.size() == 0) return false;
            }
        }
    }