Leetcode-888 两句话中的不常见单词

 1 class Solution
 2 {
 3     public:
 4         vector<string> uncommonFromSentences(string A, string B)
 5         {
 6             map<string,int> A_store,B_store;
 7             string A_words,B_words;
 8             for(int i = 0;i < A.size();i ++)
 9             {
10                 if(A[i]!=' ')
11                 {
12                     A_words += A[i];
13                 }
14                 else
15                 {
16                     A_store[A_words] ++;
17                     A_words.clear();
18                 }
19             }
20             A_store[A_words] ++;
21             A_words.clear();
22             for(int i = 0;i < B.size();i ++)
23             {
24                 if(B[i]!=' ')
25                 {
26                     B_words += B[i];
27                 }
28                 else
29                 {
30                     B_store[B_words] ++;
31                     B_words.clear();
32                 }
33             }
34             B_store[B_words] ++;
35             B_words.clear();
36             vector<string> result;
37             for(auto i = A_store.begin();i != A_store.end();i ++)
38             {
39                 string tmp = i->first;
40                 int flag = 0;
41                 for(auto j = B_store.begin();j != B_store.end();j ++)
42                 {
43                     if(j->first == tmp)
44                         flag = 1;
45                 }
46                 if(flag==1)
47                     continue;
48                 else
49                 {
50                     if(i->second==1)
51                         result.push_back(tmp);
52                 }
53             }
54             for(auto i = B_store.begin();i != B_store.end();i ++)
55             {
56                 string tmp = i->first;
57                 int flag = 0;
58                 for(auto j = A_store.begin();j != A_store.end();j ++)
59                 {
60                     if(j->first == tmp)
61                         flag = 1;
62                 }
63                 if(flag==1)
64                     continue;
65                 else
66                 {
67                     if(i->second==1)
68                         result.push_back(tmp);
69                 }
70             }
71             return result;
72         }
73 };

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转载自www.cnblogs.com/Asurudo/p/9477790.html