zoukankan      html  css  js  c++  java
  • 【leetcode】Submission Details

    Given two sentences words1, words2 (each represented as an array of strings), and a list of similar word pairs pairs, determine if two sentences are similar.
    
    For example, words1 = ["great", "acting", "skills"] and words2 = ["fine", "drama", "talent"] are similar, if the similar word pairs are pairs = [["great", "good"], ["fine", "good"], ["acting","drama"], ["skills","talent"]].
    
    Note that the similarity relation is transitive. For example, if "great" and "good" are similar, and "fine" and "good" are similar, then "great" and "fine" are similar.
    
    Similarity is also symmetric. For example, "great" and "fine" being similar is the same as "fine" and "great" being similar.
    
    Also, a word is always similar with itself. For example, the sentences words1 = ["great"], words2 = ["great"], pairs = [] are similar, even though there are no specified similar word pairs.
    
    Finally, sentences can only be similar if they have the same number of words. So a sentence like words1 = ["great"] can never be similar to words2 = ["doubleplus","good"].
    
    Note:
    
    The length of words1 and words2 will not exceed 1000.
    The length of pairs will not exceed 2000.
    The length of each pairs[i] will be 2.
    The length of each words[i] and pairs[i][j] will be in the range [1, 20].

    分析:本题要得出结果难度不大,但是会遇到Time Exceed Limited的错误,因此降低时间复杂度是关键。我的解题思路比较简单,首先为paris创建字典,找出所有与指定词有直接关系的近义词。例如paris 形成的字典如下:

    paris = [["great", "good"], ["fine", "good"], ["acting","drama"], ["skills","talent"]],
    
    dic   =  {'great': set(['good']), 'good': set(['great', 'fine']), 'talent': set(['skills']), 'skills': set(['talent']), 
    'drama': set(['acting']), 'acting': set(['drama']), 'fine': set(['good'])}

    接下来就是遍历words1和words2,遇到不相同词,直接根据key在字典中找到直接近义词,然后再利用类似广度遍历的思想,找出所有直接近义词的直接近义词,得到所有的近义词列表,再判断两个词的近义词列表是否有交集。

    class Solution(object):
        dic = {}
        def createDict(self,paris):
            self.dic = {}
            for i in paris:
                if self.dic.has_key(i[0]):
                    self.dic[i[0]].add(i[1])
                else:
                    s = set()
                    s.add(i[1])
                    self.dic[i[0]] = s
                
                if self.dic.has_key(i[1]):
                    self.dic[i[1]].add(i[0])
                else:
                    s = set()
                    s.add(i[0])
                    self.dic[i[1]] = s
        def buildWordList(self,wl):
            if len(wl) == 0:
                return ()
            s = set()
            stack = []
            for i in wl:
                stack.append(i)
            while len(stack) > 0:
                v = stack.pop()
                if v in s:
                    continue
                else:
                    s.add(v)
                    for j in self.dic[v]:
                        stack.append(j)
                            
            return s   
            
            
        def areSentencesSimilarTwo(self, words1, words2, pairs):
            """
            :type words1: List[str]
            :type words2: List[str]
            :type pairs: List[List[str]]
            :rtype: bool
            """
            if len(words1) != len(words2):
                return False
            self.createDict(pairs)
            #print self.dic
            #return
            for i in range(len(words1)):
                #print words1[i] ,words2[i]
                if words1[i] == words2[i]:
                    continue
                else:
                    s1 = set()
                    s2 = set()
                    if (self.dic.has_key(words1[i])):
                        s1 = self.buildWordList(self.dic[words1[i]])
                    if (self.dic.has_key(words2[i])):
                        s2 = self.buildWordList(self.dic[words2[i]])
                    if len(s1 & s2) == 0:
                        print s1,s2
                        print words1[i] ,words2[i]
                        return False
                        
                        
                        
            return True
  • 相关阅读:
    Muduo 网络编程示例之五: 测量两台机器的网络延迟
    “过家家”版的移动离线计费系统实现
    一种自动反射消息类型的 Google Protobuf 网络传输方案
    Muduo 设计与实现之一:Buffer 类的设计
    为什么 muduo 的 shutdown() 没有直接关闭 TCP 连接?
    Muduo 网络编程示例之八:用 Timing wheel 踢掉空闲连接
    C++ 工程实践(5):避免使用虚函数作为库的接口
    分布式系统中的进程标识
    Ormlite在一般java环境中操作Sqlite
    android游戏开发框架libgdx的使用(十八)—简单的AVG游戏效果实现
  • 原文地址:https://www.cnblogs.com/seyjs/p/7922393.html
Copyright © 2011-2022 走看看