计算连续存储的容器中特定元素出现的个数
3.归并排序 int first[] = {5,10,15,20,25}; int second[] = {50,40,30,20,10}; std::vector<int> v(10); std::sort (first,first+5); std::sort (second,second+5); std::merge (first,first+5,second,second+5,v.begin()); 4.逆序 for (int i=1; i<10; ++i) myvector.push_back(i); // 1 2 3 4 5 6 7 8 9 std::reverse(myvector.begin(),myvector.end()); // 9 8 7 6 5 4 3 2 1 5.寻找子串 for (int i=1; i<10; i++) haystack.push_back(i*10); // using default comparison: int needle1[] = {40,50,60,70}; std::vector<int>::iterator it; it = std::search (haystack.begin(), haystack.end(), needle1, needle1+4); if (it!=haystack.end()) std::cout << "needle1 found at position " << (it-haystack.begin()) << '\n'; else std::cout << "needle1 not found\n"; // using predicate comparison: int needle2[] = {20,30,50}; it = std::search (haystack.begin(), haystack.end(), needle2, needle2+3, mypredicate); if (it!=haystack.end()) std::cout << "needle2 found at position " << (it-haystack.begin()) << '\n'; else std::cout << "needle2 not found\n"; 6.排序 bool myfunction (int i,int j) { return (i<j); } struct myclass { bool operator() (int i,int j) { return (i<j);} } myobject; int myints[] = {32,71,12,45,26,80,53,33}; std::vector<int> myvector (myints, myints+8); // 32 71 12 45 26 80 53 33 // using default comparison (operator <): std::sort (myvector.begin(), myvector.begin()+4); //(12 32 45 71)26 80 53 33 // using function as comp std::sort (myvector.begin()+4, myvector.end(), myfunction); // 12 32 45 71(26 33 53 80) // using object as comp std::sort (myvector.begin(), myvector.end(), myobject); //(12 26 32 33 45 53 71 80)
