class Solution {
public:
bool areSentencesSimilar(string sentence1, string sentence2) {
}
};
1813. 句子相似性 III
一个句子是由一些单词与它们之间的单个空格组成，且句子的开头和结尾没有多余空格。比方说，"Hello World" ，"HELLO" ，"hello world hello world" 都是句子。每个单词都 只 包含大写和小写英文字母。
如果两个句子 sentence1 和 sentence2 ，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 。比方说，sentence1 = "Hello my name is Jane" 且 sentence2 = "Hello Jane" ，我们可以往 sentence2 中 "Hello" 和 "Jane" 之间插入 "my name is" 得到 sentence1 。
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 。
示例 4：
输入：sentence1 = "Luky", sentence2 = "Lucccky" 输出：false
提示：
1 <= sentence1.length, sentence2.length <= 100sentence1 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。原站题解
java 解法, 执行用时: 1 ms, 内存消耗: 40 MB, 提交时间: 2023-01-16 10:26:29
class Solution {
public boolean areSentencesSimilar(String sentence1, String sentence2) {
String[] words1 = sentence1.split(" ");
String[] words2 = sentence2.split(" ");
int i = 0, j = 0;
while (i < words1.length && i < words2.length && words1[i].equals(words2[i])) {
i++;
}
while (j < words1.length - i && j < words2.length - i && words1[words1.length - j - 1].equals(words2[words2.length - j - 1])) {
j++;
}
return i + j == Math.min(words1.length, words2.length);
}
}
javascript 解法, 执行用时: 56 ms, 内存消耗: 41 MB, 提交时间: 2023-01-16 10:26:15
/**
* @param {string} sentence1
* @param {string} sentence2
* @return {boolean}
*/
var areSentencesSimilar = function(sentence1, sentence2) {
const words1 = sentence1.split(" ");
const words2 = sentence2.split(" ");
let i = 0, j = 0;
while (i < words1.length && i < words2.length && words1[i] === words2[i]) {
i++;
}
while (j < words1.length - i && j < words2.length - i && words1[words1.length - j - 1] === words2[words2.length - j - 1]) {
j++;
}
return i + j == Math.min(words1.length, words2.length);
};
golang 解法, 执行用时: 0 ms, 内存消耗: 1.9 MB, 提交时间: 2023-01-16 10:25:48
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
}
python3 解法, 执行用时: 40 ms, 内存消耗: 15 MB, 提交时间: 2023-01-16 10:25:25
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