[LeetCode] 809. Expressive Words

Sometimes people repeat letters to represent extra feeling. For example:

• "hello" -> "heeellooo"
• "hi" -> "hiiii"

In these strings like "heeellooo", we have groups of adjacent letters that are all the same: "h", "eee", "ll", "ooo".

You are given a string s and an array of query strings words. A query word is stretchy if it can be made to be equal to s by any number of applications of the following extension operation: choose a group consisting of characters c, and add some number of characters c to the group so that the size of the group is three or more.

• For example, starting with "hello", we could do an extension on the group "o" to get "hellooo", but we cannot get "helloo" since the group "oo" has a size less than three. Also, we could do another extension like "ll" -> "lllll" to get "helllllooo". If s = "helllllooo", then the query word "hello" would be stretchy because of these two extension operations: query = "hello" -> "hellooo" -> "helllllooo" = s.

Return the number of query strings that are stretchy.

Example 1:

Input: s = "heeellooo", words = ["hello", "hi", "helo"]
Output: 1
Explanation: 
We can extend "e" and "o" in the word "hello" to get "heeellooo".
We can't extend "helo" to get "heeellooo" because the group "ll" is not size 3 or more.

Example 2:

Input: s = "zzzzzyyyyy", words = ["zzyy","zy","zyy"]
Output: 3

Constraints:

• 1 <= s.length, words.length <= 100
• 1 <= words[i].length <= 100
• s and words[i] consist of lowercase letters.

情感丰富的文字。

有时候人们会用重复写一些字母来表示额外的感受，比如 "hello" -> "heeellooo", "hi" -> "hiii"。我们将相邻字母都相同的一串字符定义为相同字母组，例如："h", "eee", "ll", "ooo"。

对于一个给定的字符串 S ，如果另一个单词能够通过将一些字母组扩张从而使其和 S 相同，我们将这个单词定义为可扩张的（stretchy）。扩张操作定义如下：选择一个字母组（包含字母 c ），然后往其中添加相同的字母 c 使其长度达到 3 或以上。

例如，以 "hello" 为例，我们可以对字母组 "o" 扩张得到 "hellooo"，但是无法以同样的方法得到 "helloo" 因为字母组 "oo" 长度小于 3。此外，我们可以进行另一种扩张 "ll" -> "lllll" 以获得 "helllllooo"。如果 s = "helllllooo"，那么查询词 "hello" 是可扩张的，因为可以对它执行这两种扩张操作使得 query = "hello" -> "hellooo" -> "helllllooo" = s。

输入一组查询单词，输出其中可扩张的单词数量。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/expressive-words
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。

这道题的思路是双指针，类似第二题。对于 words 里的每一个单词，我们拿出来和 s 做一下双指针的比较。我们每遇到一个字母的时候，我们去看一下当前这个字母最多连续出现几次。因为题目中说的可扩张的定义是某个字母在s中出现一次，就要在word中连续出现至少三次。照着这个规则，我们去判断word是否合法。

时间O(mn) - m 个单词，每个单词平均长度 n

空间O(1)

Java实现

class Solution {
    public int expressiveWords(String s, String[] words) {
        // corner case
        if (s == null || s.length() == 0 || words == null || words.length == 0) {
            return 0;
        }

        // normal case
        int count = 0;
        for (String word : words) {
            if (helper(s, word)) {
                count++;
            }
        }
        return count;
    }

    private boolean helper(String s, String word) {
        int len1 = s.length();
        int len2 = word.length();
        if (len1 < len2) {
            return false;
        }
        int i = 0, j = 0;
        while (i < len1 && j < len2) {
            char c1 = s.charAt(i);
            char c2 = word.charAt(j);
            int count1 = 0;
            int count2 = 0;
            while (i < len1 && s.charAt(i) == c1) {
                i++;
                count1++;
            }
            while (j < len2 && word.charAt(j) == c2) {
                j++;
                count2++;
            }
            if (c1 != c2 || count1 < count2 || count1 <= 2 && count1 != count2) {
                return false;
            }
        }
        return i == len1 && j == len2;
    }
}

 

LeetCode 题目总结