1.copy()函数
int myints[]={10,20,30,40,50,60,70};
std::vector<int> myvector (7);
std::copy ( myints, myints+7, myvector.begin() );
将一个容器中的元素复制到另一个容器中
2.count()函数
int myints[] = {10,20,30,30,20,10,10,20}; // 8 elements
int mycount = std::count (myints, myints+8, 10);
std::cout << "10 appears " << mycount << " times.\n";
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)
STL常用算法
原文：http://blog.csdn.net/yutianxin123/article/details/52210096
