STL对左旋转字符串针对三种不同数据结构进行了不同的实现。
对单向链表采用的是同步位移，双向链表是三次翻转，都很简单，主要看看针对随机存取的数组做的循环位移实现。
STL这个版本的源码如下：
template <class RandomAccessIterator, class Distance>void __rotate(RandomAccessIterator first, RandomAccessIterator middle, RandomAccessIterator last, Distance*, random_access_iterator_tag){// gcd是求最大公约数的函数。Distance n = __gcd(last – first, middle – first);while (n–)// 需要执行__rotate_cycle n次。__rotate_cycle(first, last, first + n, middle – first, value_type(first));}template <class RandomAccessIterator, class Distance, class T>void __rotate_cycle(RandomAccessIterator first, RandomAccessIterator last, RandomAccessIterator initial, Distance shift, T*){T value = *initial;RandomAccessIterator ptr1 = initial;RandomAccessIterator ptr2 = ptr1 + shift;while (ptr2 != initial) {*ptr1 = *ptr2;ptr1 = ptr2;if (last – ptr2 > shift)ptr2 += shift;elseptr2 = first + (shift – (last – ptr2));}*ptr1 = value;}template <class EuclideanRingElement>EuclideanRingElement __gcd(EuclideanRingElement m, EuclideanRingElement n){while (n != 0) {EuclideanRingElement t = m % n;m = n;n = t;}return m;}__gcd是求两个数的最大公约数，也是循环位移的遍数。
举个例子来说明算法过程，数组123456789，把123翻转到右边，*first=1,*last=9,*middle=4;
要旋转字符串（123）的长度为3，字符串长度为9，3和9的最大公约数为3，因此需要翻转3遍；
第一遍从*(initial+shift)=6开始，6移到3的位置，9移到6的位置，下一个位置是ptr2 = first + (shift – (last – ptr2))=0+(3-(8-8))=3，不满足ptr2 != initial的条件，退出循环，，然后*ptr1 = value，即把数字3移动到数字9的位置，从而完成了3,6,9三个数字的位移，下面的2遍循环则分别完成2,5,8和1,4,76个数字的位移，最后得到最终结果456789123。
整个算法过程可用下图表示：
自我实现C++的int版：
#include <iostream>using namespace std;int __gcd(int a, int b){while (b){int tmp = a%b;a = b;b = tmp;}return a;}void __rotate_cycle(int *data, int first, int last, int initial, int shift){int val = data[initial];int pos1 = initial;int pos2 = initial + shift;while (pos2 != initial){data[pos1] = data[pos2];pos1 = pos2;pos2 = (pos2 + shift) % (last – first + 1);}data[pos1] = val;}void __rotate(int *data,int first, int middle, int last){int gcd = __gcd(last – first+1, middle – first);cout << "循环位移" << gcd << "遍" << endl;while (gcd–){__rotate_cycle(data, first, last, first + gcd, middle – first);}}void main(){int data1[10] = {1,2,3,4,5,6,7,8,9,10};__rotate(data1, 0, 3, 9);for (int i = 0; i < 10; i++){cout << data1[i] << " ";}cout << endl;int data2[10] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };__rotate(data2, 0, 4, 9);for (int i = 0; i < 10; i++){cout << data2[i] << " ";}cout << endl;int data3[10] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };__rotate(data3, 0, 5, 9);for (int i = 0; i < 10; i++){cout << data3[i] << " ";}cout << endl;int data4[10] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };__rotate(data4, 0, 6, 9);for (int i = 0; i < 10; i++){cout << data4[i] << " ";}int c = 0;cin >> c;}运行结果：
那里面非常漂亮，个个观景区都能看到奇形怪状的岩石。
