9/20 每日一題(找出生僻字)

 A sentence is a string of single-space separated words where each word consists only of lowercase letters.

A word is uncommon if it appears exactly once in one of the sentences, and does not appear in the other sentence.

Given two sentences s1 and s2, return a list of all the uncommon words. You may return the answer in any order.

 

Example 1:

Input: s1 = "this apple is sweet", s2 = "this apple is sour"
Output: ["sweet","sour"]

Example 2:

Input: s1 = "apple apple", s2 = "banana"
Output: ["banana"]

 

Constraints:

• 1 <= s1.length, s2.length <= 200
• s1 and s2 consist of lowercase English letters and spaces.
• s1 and s2 do not have leading or trailing spaces.
• All the words in s1 and s2 are separated by a single space.

class Solution:

    def uncommonFromSentences(self, s1: str, s2: str) -> List[str]:

        s_list=s1.split()

        s_list.extend(s2.split())

        print(s_list)

        s_dict=Counter(s_list)

       

        print(s_dict)

        res=[]

        for k,v in s_dict.items():

            if v==1:

                res.append(k)

        print(res)

        return res

-----------------------------

from collections import Counter

class Solution:
    def getList(self, s: str) -> (List[str], List[str]):
        res = Counter(s)
        tem1 = set()
        tem2 = set()
        for i in res:
            if res[i] == 1:
                tem1.add(i)
            else:
                tem2.add(i)
        return (tem1, tem2)

    def uncommonFromSentences(self, s1: str, s2: str) -> List[str]:
        if not s1: return s2
        if not s2: return s1
        
        l1, r1 = self.getList(s1.split(" "))
        l2, r2 = self.getList(s2.split(" "))

        return (l1^l2) - r1 - r2

       

  ----------------------------------------

class Solution:
    def uncommonFromSentences(self, s1: str, s2: str) -> List[str]:
        l1=s1.split()
        l2=s2.split()
        l3=[]
        for i in range(0,len(l1)):
            if l1[i] not in l2:
                if l1.count(l1[i])==1:
                    l3.append(l1[i])
        for i in range(0,len(l2)):
            if l2[i] not in l1:
                if l2.count(l2[i])==1:
                    l3.append(l2[i])

return l3