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 <= 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。
class Solution:
def areSentencesSimilar(self, sentence1: str, sentence2: str) -> bool:
if sentence1 == sentence2:
return True
n1, n2 = len(sentence1), len(sentence2)
if n1 == n2:
return False
if n1 < n2:
sentence1, sentence2 = sentence2, sentence1
words1, words2 = sentence1.split(), sentence2.split()
i = j = 0
while i < len(words2) and words1[i] == words2[i]:
i += 1
if i == len(words2):
return True
while j < len(words2) and words1[len(words1) - 1 - j] == words2[len(words2) - 1 - j]:
j += 1
if j == len(words2):
return True
return i + j == len(words2)
class Solution {
public boolean areSentencesSimilar(String sentence1, String sentence2) {
if (Objects.equals(sentence1, sentence2)) {
return true;
}
int n1 = sentence1.length(), n2 = sentence2.length();
if (n1 == n2) {
return false;
}
if (n1 < n2) {
String t = sentence1;
sentence1 = sentence2;
sentence2 = t;
}
String[] words1 = sentence1.split(" ");
String[] words2 = sentence2.split(" ");
int i = 0, j = 0;
while (i < words2.length && Objects.equals(words1[i], words2[i])) {
++i;
}
if (i == words2.length) {
return true;
}
while (j < words2.length && Objects.equals(words1[words1.length - 1 - j], words2[words2.length - 1 - j])) {
++j;
}
if (j == words2.length) {
return true;
}
return i + j == words2.length;
}
}