2017-11 (not fresh grad ) find all substring with N size and only one duplicate character.

可能是2018 亚麻的OA题。

1.给字符串， 找出里面长度为N的， 并且里面字符只有一次重复的字串。
例子：s: asdfaghjkjqoiiii;N=5.  返回asdfa ghjkj hjkjq jkjqo jqoii.

 1 #include <iostream>     // std::cout
 2 #include <algorithm>    // std::make_heap, std::pop_heap, std::push_heap, std::sort_heap
 3 #include <vector>       // std::vector
 4 #include <unordered_map>
 5 #include <unordered_set>
 6 #include <numeric>
 7 #include <sys/time.h>
 8 #include <list>
 9 #include <map>
10
11 //main point : ( refer to key data structure and the trick  )
12
13 // use the two unordered set: dup and match
14 //  dup : for check if the finded substring is already in result set , that means it has been found early.
15 //  match: check if the substring has only one duplicated char.
16
17 //use list to store the results.
18
19 // iterate the string with N size window from first char to the end of the string.
20
21
22 using namespace std;
23
24
25 list<string> findNSubstring (string& s, int N){
26     //check the input
27     if ( s.empty()|| s.size() < N ) return list<string>();
28
29     int sp =0;
30     int steps = N-1;
31     cout << " the steps is " << steps << endl;
32     unordered_set<string> dup;  //make sure no duplicated substring in result set.
33     unordered_set <char> match;   // check the substring has only one duplicated char.
34     list<string> res;    //store the output.
35
36     // Check the substrings one by one with N size window.
37     while ((sp+steps) < s.size()){
38         match.clear();
39
40         // check if the substring have only one duplicated character.
41         for (int j =0; j < N ; j++){
42             match.insert(s[sp+j]);
43         }
44
45         if (match.size() == N-1){ // if yes, check if that substring has been in result set.
46             cout<< "find one " <<  s.substr( sp, N)  << " at "<< sp << endl;
47             if ( dup.find( s.substr( sp, steps)) == dup.end()){
48                 res.push_back(s.substr( sp, N));
49                 dup.insert (s.substr( sp, N));
50             }
51         }
52         sp++;
53     }
54     return res;
55 }
56
57 int main () {
58
59    // cout << " test  a substring length : " << string ("").size() << endl;
60
61     //get the start time.
62     struct timeval tv;
63     gettimeofday(&tv,NULL);
64     long ts = tv.tv_sec * 1000 + tv.tv_usec / 1000;
65
66     //*** call the function .
67
68     string in = "iiiiiiiiiiiiiiiiiii";//asdfaghjkjqoiiii";
69
70     int q = 100 ;
71
72     auto out = findNSubstring(in, q) ;
73
74     for (auto e : out)
75         cout<< e << endl;
76     //*** end of the call
77
78     //get the time of end.
79     gettimeofday(&tv,NULL);
80     long te = tv.tv_sec * 1000 + tv.tv_usec / 1000;
81
82     //output the time of the running.
83     cout<<endl<< endl<< "running tmie is : " << te - ts << endl;
84     return 0;
85 }

原文地址：https://www.cnblogs.com/HisonSanDiego/p/8295998.html