《编程珠玑》第二章，三种字符串旋转算法

▶ 将长度为 len 的字符串旋转 dist 位的意思是：将该字符串的首字母放到该字符串末尾，重复该操作 dist 次

● 代码

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>

int gcd(int a, int b)                           // 求最大公约数
{
    a = (a > b ? a : b);
    b = (a > b ? b : a);
    for(int temp = a % b;temp != 0; temp = a % b)
    {
        a = b;
        b = temp;        
    }
    return b;
}

int spin1(char *ss, const int len, int dist)    // 跳跃下标法
{    
    for (int i = 0, iend = gcd(len, dist); i < iend; ++i)
    {
        char temp = ss[i];
        int oldIndex = i;
        for (int newIndex = (oldIndex + dist) % len; newIndex != i; newIndex = (oldIndex + dist) % len)
        {
            ss[oldIndex] = ss[newIndex];
            oldIndex = newIndex;
        }
        ss[oldIndex] = temp;
    }
    return 0;
}

int swapTile(char *a, char *b, int tileLenght)  // 交换 a 和 b 的前 tileLength 位
{
    for (int i = 0; i < tileLenght; ++i)
    {
        char temp = a[i];
        a[i] = b[i];
        b[i] = temp;
    }
    return 0;
}

int spin2(char *ss, const int len, int dist)    // 分块交换法
{
    if (dist%len == 0)
        return 0;

    int i = dist, p = dist;
    for (int j = len - dist; i != j;)
    {
        if (i > j)
        {
            swapTile(ss + p - i, ss + p, j);
            i -= j;
        }
        else
        {
            swapTile(ss + p - i, ss + p + j - i, i);
            j -= i;
        }
    }
    swapTile(ss + p - i, ss + p, i);
    return 0;
}

int spin3(char *ss, const int len, int dist)    // 求逆法
{
    int distR = dist % len;
    if (distR == 0)
        return 0;
    for (int i = 0, j = distR - 1; i < j; ++i, --j)     // 翻左手
    {
        char temp = ss[i];
        ss[i] = ss[j];
        ss[j] = temp;
    }
    for (int i = distR, j = len - 1; i < j; ++i, --j)   // 翻右手
    {
        char temp = ss[i];
        ss[i] = ss[j];
        ss[j] = temp;
    }
    for (int i = 0, j = len - 1; i < j; ++i, --j)       // 两手一起翻
    {
        char temp = ss[i];
        ss[i] = ss[j];
        ss[j] = temp;
    }
    return 0;
}

int testGcd(const int target)                   // 测试求最大公约数的方法
{
    for (int i = 1; i <= target; ++i)
    {
        int temp1 = gcd(target, i), temp2 = gcd(target / temp1, i / temp1);
        printf("(%4d,%4d) = %4d, (%4d,%4d) = %4d
", target, i, temp1, target / temp1, i / temp1, temp2);
    }
    return 0;
}

int testSpinSmall()                             // 测试旋转正确性
{
    char origin[] = "abcdefghijklmnopqrstuvwxyz0123456789!";
    char temp[] = "abcdefghijklmnopqrstuvwxyz0123456789!";
    
    printf("Test spin 1:
%4d->%s
", 0, origin);
    for (int i = 1; i < 100; ++i)    
    {
        strcpy(temp, origin);
        spin1(temp, strlen(origin), i);
        printf("%4d->%s
", i, temp);
    }
    printf("Test spin 2:
%4d->%s
", 0, origin);
    for (int i = 1; i < 100; ++i)
    {
        strcpy(temp, origin);
        spin2(temp, strlen(origin), i);
        printf("%4d->%s
", i, temp);
    }

    printf("Test spin 3:
%4d->%s
", 0, origin);
    for (int i = 1; i < 100; ++i)
    {
        strcpy(temp, origin);
        spin3(temp, strlen(origin), i);
        printf("%4d->%s
", i, temp);
    }
    return 0;
}

int testSpinLarge()                             // 测试旋转效率，对长为 size 的数组旋转指定位数 time 次
{
    const int size = 10000000, time = 10;
    char *origin = (char *)malloc(sizeof(char)*size);    
    clock_t t1, t2, tE;    
    for (int count = 1; count <= 100; count++)
    {
        t1 = clock();
        for (int i = 1; i < time; ++i)
            spin3(origin, size, i);
        t2 = clock();
        printf("%4d->%10d ms
", count, t2 - t1);
    }      
    return 0;
}

int main()
{
    testGcd(24);    
    testSpinSmall();
    testSpinLarge();
    getchar();
    return 0;
}

● 测试结果，在长度为 1000000 的字符串上分别旋转 1~100 位各 10 次，记录个算法消耗的时间