面试题：一千万条短信中，找到重复次数最多的10条 

  1 #include<iostream>
  2 #include<fstream>
  3 #include <string>
  4 #include <unordered_map>
  5 #include <vector>
  6 #include <algorithm>
  7 #include <queue>
  8 
  9 using namespace std;
 10 
 11 typedef unsigned long int uint64_t;
 12 
 13 //获取字符串的hash值
 14 uint64_t getHash(const string& s) {
 15     uint64_t hashVal = 0;
 16     uint64_t a = 3589;
 17     uint64_t b = 81369;
 18     for (char ch : s) {
 19         hashVal = hashVal * a + ch;
 20         a = a * b;
 21     }
 22     return hashVal;
 23 }
 24 
 25 struct posAndCount {
 26     int pos;
 27     int count;
 28     posAndCount(int a = 0, int b = 0) :
 29             pos(a), count(b) {
 30     };
 31 };
 32 
 33 struct NameHashPosAndCount {
 34     uint64_t namehash;
 35     int pos;
 36     int count;
 37 
 38     friend bool operator <(const NameHashPosAndCount& a,
 39             const NameHashPosAndCount& b) {
 40         return a.count > b.count;
 41     }
 42 };
 43 
 44 
 45 //从N个数据中找出出现次数最多的N个
 46 void MfromN() {
 47     unordered_map<uint64_t, posAndCount> names;
 48     unordered_map<uint64_t, posAndCount>::iterator it;
 50     posAndCount pc;
 51     int pos = 0;

 52     ifstream fin("message.txt");
 53     string l;
 54     while (fin >> l) {
 55         uint64_t hashVal = getHash(l);
 56         it = names.find(hashVal);
 57         if (it == names.end()) {
 58             pc.pos = pos;
 59             pc.count = 0;
 60             names[hashVal] = pc;
 61         } else {
 62             names[hashVal].count++;
 63         }
 64         pos++;
 65     }
 66     fin.close();
 67 
 68 
 69     //优先队列寻找出现次数最多的10条
 70     priority_queue<NameHashPosAndCount> name10;
 71     NameHashPosAndCount n;
 72     for (it = names.begin(); it != names.end(); it++) {
 73         if (name10.size() < 10) {
 74             n.namehash = it->first;
 75             n.pos = it->second.pos;
 76             n.count = it->second.count;
 77             name10.push(n);
 78         } else {
 79             if (name10.top().count < it->second.count) {
 80                 name10.pop();
 81                 n.namehash = it->first;
 82                 n.pos = it->second.pos;
 83                 n.count = it->second.count;
 84                 name10.push(n);
 85             }
 86         }
 87     }
 88     vector<int> index;
 89     while (!name10.empty()) {
 90         index.push_back(name10.top().pos);
 91         name10.pop();
 92     }
 93     sort(index.begin(),index.end());
 94 
 95     ifstream fin2("message.txt");
 96     int i = 0;
 97     int k = 0;
 98     while(k < 10 && fin2 >> l){
 99         if(i == index[k]){
100             cout << l << endl;
101             k++;
102         }
103         i++;
104     }
105     fin2.close();
106 }

 

posted @ 2017-02-28 14:38  wxquare  阅读(899)  评论(0)    收藏  举报