25.不改变原生数据的STL algorithm

Posted 2020-10-28 喵小喵~

• 通过仿函数for_each操作

  1 vector<int> myv{ 1,2,3,4,5 };
  2 list<double> db{ 1.1,2.2,3.3,4.4,5.5 };
  3 
  4 //循环算法,算法的泛型
  5 print p = for_each(db.begin(), db.end(), print());
  6 cout << p.count << endl;

   

• find_if查找算法

  1 //查找算法
  2     auto it = find(myv.begin(), myv.end(), 3);
  3     cout << *it << endl;

   

• find_if与lambda结合

   1 //寻找第一个偶数的位置,返回0就是找到的
   2     auto it = find_if(myv.begin(), myv.end(), [](int x)->bool 
   3     {
   4         if (x % 2 == 0)
   5         {
   6             return 0;
   7         }
   8         else
   9         {
  10             return 1;
  11         }
  12     });
  13 
  14     if (it != myv.end())
  15     {
  16         cout << *it << "  pos=" << it - myv.begin() << endl;
  17     }

   

• adjacent_find处理相邻的两个数据

  1 list <int> mylist{ 3,6,9,11,11,28,20,29 };
  2     //查找相邻的两个元素相等的位置(adjacent_find处理相邻的两个数据)
  3     auto it = adjacent_find(mylist.begin(), mylist.end());
  4     if (it != mylist.end())
  5     {
  6         cout << *it << endl;
  7         it++;
  8         cout << *it << endl;
  9     }

   

• adjacent_find与lambda结合

   1 //寻找第一个奇偶性不同的
   2         auto it = adjacent_find(mylist.begin(), mylist.end(), [](int x,int y)->bool
   3         {
   4             if ((x - y) % 2 == 0)
   5             {
   6                 return 0;
   7             }
   8             else
   9             {
  10                 return 1;
  11             }
  12         });
  13         if (it != mylist.end())
  14         {
  15             cout << *it << endl;
  16             it++;
  17             cout << *it << endl;
  18         }

   

• find_first_of寻找第一个集合在第二个集合中出现的第一个数据

   1 char *str1 = "1234567890";
   2     char *str2 = "abcdefg123";
   3     //寻找在string1中第一个出现在string2的字符
   4     char *p = find_first_of(str1, str1 + strlen(str1), str2, str2 + strlen(str2));
   5     cout << *p << endl;
   6 
   7     vector<int> myint1{ 1,2,3,4,5 };
   8     vector<int> myint2{ 7,8,9,10,1 };
   9     auto it = find_first_of(myint1.begin(), myint1.end(), myint2.begin(), myint2.end());
  10     cout << *it << endl;

   

• count与count_if查询数据个数

   1 vector<int> myint{ 1,2,3,4,5,6,1,2,3,4 };
   2     int count1 = count(myint.begin(), myint.end(), 3);
   3     cout << count1 << endl;
   4 
   5     //查询所有小于4的元素的个数
   6     int count2 = count_if(myint.begin(), myint.end(), [](int x)
   7     {
   8         if (x < 4)
   9         {
  10             return 1;
  11         }
  12         else
  13         {
  14             return 0;
  15         }
  16     });
  17     cout << count2 << endl;

   

• mismatch判断两个集合是否相等

   1 vector<int> myint1{ 1,2,3,4,5,6,1,2,3,4 };
   2     vector<int> myint2{ 1,2,3,4,5,6,1,2,3,4 };
   3 
   4     auto it = mismatch(myint1.begin(), myint1.end(), myint2.begin());
   5     if (it.first == myint1.end() && it.second == myint2.end())
   6     {
   7         cout << "相等" << endl;
   8     }
   9     else
  10     {
  11         //first是第一个容器不匹配的位置(可能为空),second是第二个容器不匹配的位置(可能为空)
  12         cout << "不相等" << endl;
  13         cout << *(it.first) <<"  " << *(it.second) << endl;
  14     }

   1 char *s1[] = { "abc","acv","adf","oop" };
   2     char *s2[] = { "abc","acv","adf","oop","134" };
   3     auto it = mismatch(s1, s1 + 4, s2, [](const char* str1,const char* str2)
   4     {
   5         if (strcmp(str1, str2) == 0)
   6         {
   7             return 1;
   8         }
   9         else
  10         {
  11             return 0;
  12         }
  13     });
  14 
  15     if (it.first == s1 + sizeof(s1)/sizeof(s1[0]) && it.second == s2 + sizeof(s2) / sizeof(s2[0]))
  16     {
  17         cout << "相等" << endl;
  18     }
  19     else
  20     {
  21         //first是第一个容器不匹配的位置(可能为空),second是第二个容器不匹配的位置(可能为空)
  22         cout << "不相等" << endl;
  23         //cout << *(it.first) << "  " << *(it.second) << endl;
  24     } 

   

• equal判断是否是一样的集合

   1     
   2 
   3     vector<int> myv1{ 1,2,3,4,5 };
   4     vector<int> myv2{ -1,2,-3,4,-5 };
   5 
   6     //判断绝对值知否相等
   7     if (equal(myv1.begin(), myv1.end(), myv2.begin(), [](int a, int b) ->bool
   8     {
   9         if (a == abs(b) || b == abs(a))
  10         {
  11             return 1;
  12         }
  13         else
  14         {
  15             return 0;
  16         }
  17     }))
  18     {
  19         cout << "相等" << endl;
  20     }
  21     else
  22     {
  23         cout << "不相等" << endl;
  24     }

   

• search判断有没有连续一样的部分

   1 vector<int> myv1{ 1,2 };
   2     vector<int > myv2{ 1,2,3,4,5 };
   3     
   4     //判断有没有连续相等的子集
   5     auto it = search(myv1.begin(), myv1.end(), myv2.begin(), myv2.end());
   6 
   7     if (it == myv1.end())
   8     {
   9         cout << "是子集" << endl;
  10     }
  11     else
  12     {
  13         cout << "不是子集" << endl;
  14     }

   

• search_n判断有没有连续一样的数据

   1 vector<int> myv1{ 1,2,2,2,3,3,4 };
   2     
   3     //判断有没有连续相等的数据
   4     auto it = search_n(myv1.begin(), myv1.end(),3, 2);
   5     
   6     if (it == myv1.end())
   7     {
   8         cout << "有" << endl;
   9     }
  10     else
  11     {
  12         cout << "没有" << endl;
  13     }

   

• 从反向寻找一个集合在另一个集合中出现的位置

  1 vector<int> myv2{ 1,2,2,2,3,3,4 };
  2     vector<int> myv3{ 3,4 };
  3     //在myv2中从后往前找myv3所在的位置
  4     auto it = find_end(myv2.begin(), myv2.end(), myv3.begin(), myv3.end());
  5 
  6     if(it != myv2.end())
  7     {
  8         cout << *it << "   " << it - myv2.begin() << endl;
  9     }