给你一个字符串句子sentence和检索词word，其中句子由若干空格分隔的单词组成。请编写一个函数，判断检索词word是否是句子中任意单词的前缀。
如果是句子中某个单词的前缀，则返回该单词出现的下标（从1开始）；如果存在匹配多个，则取第一个单词的下标；如果不存在，则返回-1。
示例：
输入：sentence = “i love hogwarts”, word = “hog”
输出：3
解释："hog"是单词"hogwarts"的前缀，而"hogwarts"是第3个单词。
题目难度：简单
题目来源：力扣 1455
def solution(sentence: str, word: str) -> int:
# your code here
assert solution("i love hogwarts") == 3
assert solution("this problem is an easy problem", "pro") == 2
def solution(sentence: str, word: str) -> int:
# 0309 单词前缀
sen_list = sentence.split(" ")
for sen in sen_list:
if sen.startswith(word):
return sen_list.index(sen)+1
return -1
assert solution("i love hogwarts", 'hog') == 3
assert solution("this problem is an easy problem", "pro") == 2
return 0 if len([a for a in str(sentence).split(" “) if a.find(word)==0])==0 else int([xx for xx in range(len([b for b in str(sentence).split(” “) ])) if [b for b in str(sentence).split(” “) ][xx]==[a for a in str(sentence).split(” ") if a.find(word)==0][0]][0])+1
def solution(sentence: str, word: str) -> int:
words = sentence.split(" ")
for i in words:
if word in i and word[0] == i[0]:
return words.index(i) + 1
else:
return -1
assert solution("i love hogwarts", "hog") == 3
assert solution("this problem is an easy problem", "pro") == 2
public int isPrefixOfWord(String sentence, String searchWord) {
// 题解一
// String[] strArray = sentence.split(" ");
// for(int i=0;i<strArray.length;i++){
// if(strArray[i].startsWith(searchWord))
// return i+1;
// }
// return -1;
// 题解二
// 记录第几个单词
int index = 1;
// 记录searchWord的第几个字母
int indexSearch = 0,len = searchWord.length()-1;
// 记录是否以searchWord开头
boolean isStart = true;
for(int i=0;i<sentence.length();i++){
if(sentence.charAt(i) == (' ')){
index++;
indexSearch = 0;
isStart = true;
}else{
if(indexSearch <= len){
if(sentence.charAt(i) != searchWord.charAt(indexSearch)){
// 只要有不一样的就置为false
isStart = false;
}
if(indexSearch++ == len && isStart)
return index;
}
}
}
return -1;
} def solution(sentence: str, word: str) -> int:
for k,v in enumerate(sentence.split(' ')):
if v.startswith(word):
return k + 1
return -1
def solution(sentence: str, word: str) -> int:
# your code here
list_words = sentence.split(" ")
for i in range(len(list_words)):
if word == list_words[i][0:len(word)]:
return i + 1
return -1
assert solution("i love hogwarts", "hog") == 3
assert solution("i love hogwarts", "s") == -1
assert solution("this problem is an easy problem", "pro") == 2def solution(sentence: str, word: str) -> int:
# your code here
list1 = sentence.split(" ")
for s in list1:
if s.startswith(word):
return list1.index(s)+1
return -1
def solution(sentence: str, word: str):
li=sentence.split(" ")
for i in range(len(li)):
if word==li[i][:3]:
return i+1
return -1
assert solution("i love hogwarts","hog") == 3
assert solution("this problem is an easy problem", "pro") == 2
def solution(sentence: str, word: str) -> int:
list1 = sentence.split()
for i in list1:
if word in i:
return list1.index(i)+1
assert solution("i love hogwarts", "hog") == 3
assert solution("this problem is an easy problem", "pro") == 2
def solution(sentence: str, word: str) -> int:
lis = sentence.split(' ')
for i in range(len(lis)):
if lis[i][:len(word)] == word:
return i+1
return -1
assert solution("i love hogwarts","hogwarts") == 3
assert solution("this problem is an easy problem", "pro") == 2
def get_min_index(sentence: str, word: str) -> int:
# your code here
str_list = sentence.split(" ")
resoult_int = -1
index = 1
for i in str_list:
if i[0:len(word)] in word:
resoult_int = index
break
index +=1
return resoult_int
def solution(sentence: str, word: str) -> int:
a=[i for i in sentence.split() if i[:len(word)] == word]
return sentence.split().index(a[0])+1 if len(a)>0 else -1
def solution(sentence: str, word: str) -> int:
#方法一
#result =[index for index,value in enumerate(sentence.split(' ')) if value.startswith(word)]
#return -1 if len(result) == 0 else result[0]+1
#方法二
for index, value in enumerate(sentence.split(' ')):
if value.startswith(word):
return index+1
return -1