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1813. 句子相似性 III

题目描述

给定两个字符串 sentence1 和 sentence2，每个表示由一些单词组成的一个句子。句子是一系列由 单个 空格分隔的 单词，且开头和结尾没有多余空格。每个单词都只包含大写和小写英文字母。

如果两个句子 s1 和 s2 ，可以通过往其中一个句子插入一个任意的句子（可以是空句子）而得到另一个句子，那么我们称这两个句子是 相似的 。注意，插入的句子必须与现有单词用空白隔开。 

比方说，

  • s1 = "Hello Jane" 与 s2 = "Hello my name is Jane"，我们可以往 s1 中 "Hello" 和 "Jane" 之间插入 "my name is" 得到 s2 。
  • s1 = "Frog cool" 与 s2 = "Frogs are cool" 不是相似的，因为尽管往 s1 中插入 "s are"，它没有与 "Frog" 用空格隔开。

给你两个句子 sentence1 和 sentence2 ，如果 sentence1 和 sentence2 是 相似 的，请你返回 true ，否则返回 false 。

 

示例 1：

输入：sentence1 = "My name is Haley", sentence2 = "My Haley"
输出：true
解释：可以往 sentence2 中 "My" 和 "Haley" 之间插入 "name is" ，得到 sentence1 。
 

示例 2：

输入：sentence1 = "of", sentence2 = "A lot of words"
输出：false
解释：没法往这两个句子中的一个句子只插入一个句子就得到另一个句子。
 

示例 3：

输入：sentence1 = "Eating right now", sentence2 = "Eating"
输出：true
解释：可以往 sentence2 的结尾插入 "right now" 得到 sentence1 。

 

提示：

  • 1 <= sentence1.length, sentence2.length <= 100
  • sentence1 和 sentence2 都只包含大小写英文字母和空格。
  • sentence1 和 sentence2 中的单词都只由单个空格隔开。

解法

方法一：双指针

我们将两个句子按照空格分割成两个单词数组 words1 和 words2，假设 words1 和 words2 的长度分别为 $m$ 和 $n$，不妨设 $m \geq n$。

我们使用双指针 $i$ 和 $j$，初始时 $i = j = 0$。接下来，我们循环判断 words1[i] 是否等于 words2[i]，是则指针 $i$ 继续右移；然后我们循环判断 words1[m - 1 - j] 是否等于 words2[n - 1 - j]，是则指针 $j$ 继续右移。

循环结束后，如果 $i + j \geq n$，说明两个句子相似，返回 true，否则返回 false。

时间复杂度 $O(L)$，空间复杂度 $O(L)$。其中 $L$ 为两个句子的长度之和。

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class Solution:
    def areSentencesSimilar(self, sentence1: str, sentence2: str) -> bool:
        words1, words2 = sentence1.split(), sentence2.split()
        m, n = len(words1), len(words2)
        if m < n:
            words1, words2 = words2, words1
            m, n = n, m
        i = j = 0
        while i < n and words1[i] == words2[i]:
            i += 1
        while j < n and words1[m - 1 - j] == words2[n - 1 - j]:
            j += 1
        return i + j >= n

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class Solution {
    public boolean areSentencesSimilar(String sentence1, String sentence2) {
        var words1 = sentence1.split(" ");
        var words2 = sentence2.split(" ");
        if (words1.length < words2.length) {
            var t = words1;
            words1 = words2;
            words2 = t;
        }
        int m = words1.length, n = words2.length;
        int i = 0, j = 0;
        while (i < n && words1[i].equals(words2[i])) {
            ++i;
        }
        while (j < n && words1[m - 1 - j].equals(words2[n - 1 - j])) {
            ++j;
        }
        return i + j >= n;
    }
}

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class Solution {
public:
    bool areSentencesSimilar(string sentence1, string sentence2) {
        auto words1 = split(sentence1, ' ');
        auto words2 = split(sentence2, ' ');
        if (words1.size() < words2.size()) {
            swap(words1, words2);
        }
        int m = words1.size(), n = words2.size();
        int i = 0, j = 0;
        while (i < n && words1[i] == words2[i]) {
            ++i;
        }
        while (j < n && words1[m - 1 - j] == words2[n - 1 - j]) {
            ++j;
        }
        return i + j >= n;
    }

    vector<string> split(string& s, char delim) {
        stringstream ss(s);
        string item;
        vector<string> res;
        while (getline(ss, item, delim)) {
            res.emplace_back(item);
        }
        return res;
    }
};

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func areSentencesSimilar(sentence1 string, sentence2 string) bool {
    words1, words2 := strings.Fields(sentence1), strings.Fields(sentence2)
    if len(words1) < len(words2) {
        words1, words2 = words2, words1
    }
    m, n := len(words1), len(words2)
    i, j := 0, 0
    for i < n && words1[i] == words2[i] {
        i++
    }
    for j < n && words1[m-1-j] == words2[n-1-j] {
        j++
    }
    return i+j >= n
}

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function areSentencesSimilar(sentence1: string, sentence2: string): boolean {
    const [words1, words2] = [sentence1.split(' '), sentence2.split(' ')];
    const [m, n] = [words1.length, words2.length];

    if (m > n) return areSentencesSimilar(sentence2, sentence1);

    let [l, r] = [0, 0];
    for (let i = 0; i < n; i++) {
        if (l === i && words1[i] === words2[i]) l++;
        if (r === i && words2[n - i - 1] === words1[m - r - 1]) r++;
    }

    return l + r >= m;
}

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function areSentencesSimilar(sentence1, sentence2) {
    const [words1, words2] = [sentence1.split(' '), sentence2.split(' ')];
    const [m, n] = [words1.length, words2.length];

    if (m > n) return areSentencesSimilar(sentence2, sentence1);

    let [l, r] = [0, 0];
    for (let i = 0; i < n; i++) {
        if (l === i && words1[i] === words2[i]) l++;
        if (r === i && words2[n - i - 1] === words1[m - r - 1]) r++;
    }

    return l + r >= m;
}

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