题目如下:
Given a list of pairs of equivalent words
synonyms
and a sentencetext
, Return all possible synonymous sentences sorted lexicographically.Example 1:
Input: synonyms = [["happy","joy"],["sad","sorrow"],["joy","cheerful"]], text = "I am happy today but was sad yesterday" Output: ["I am cheerful today but was sad yesterday", "I am cheerful today but was sorrow yesterday", "I am happy today but was sad yesterday", "I am happy today but was sorrow yesterday", "I am joy today but was sad yesterday", "I am joy today but was sorrow yesterday"]Constraints:
0 <= synonyms.length <= 10
synonyms[i].length == 2
synonyms[0] != synonyms[1]
- All words consist of at most
10
English letters only.text
is a single space separated sentence of at most10
words.
解题思路:我的方法是替换,依次遍历synoyms,如果text中存在synoyms[i][0],把text中的第一个synoyms[i][0]替换成synoyms[i][1];同样,如果存在synoyms[i][1],把text中的第一个synoyms[i][1]替换成synoyms[i][0]。然后再对替换过的text做同样的操作,直到不能替换位置。考虑到synoyms[i][0] = 'a',而text中存在'an'这种情况,为了防止误替换,可以把每个单词的前后都加上'#'作为分隔。
代码如下:
class Solution(object): def generateSentences(self, synonyms, text): """ :type synonyms: List[List[str]] :type text: str :rtype: List[str] """ text = '#' + text.replace(' ', '#') + '#' queue = [text] for (w1,w2) in synonyms: for text in queue : newtext = text.replace('#' + w1+'#','#' + w2+'#',1) if newtext != text and newtext not in queue: queue.append(newtext) newtext = text.replace('#' + w2 + '#', '#' + w1 + '#',1) if newtext != text and newtext not in queue: queue.append(newtext) res = [] for i in sorted(queue): newtext = i.replace('#', ' ') res.append(newtext[1:-1]) return res