字符串的旋转

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给定一个字符串abcdef,将前3个移到后面变成defabc.

#include <iostream>using namespace  std;//解法1:暴力/*void LeftshiftOne(char *s,int n){    char a = s[0];    for (int i = 0; i < n;i++)    {        s[i] = s[i + 1];    }    s[n-1] = a;}void LeftRotateString(char *s,int m,int n){    while (m--)    {        LeftshiftOne(s, n);    }}int main(){    char s[100];    cin >> s;    LeftRotateString(s,3,strlen(s));    cout << s;}//时间复杂度过高o(m*n),空间复杂度O(1);*/

解法2:三步反转

void ReverseString(char *s,int from,int to){    while (from<to)    {        char t = s[from];        s[from++] = s[to];        s[to--] = t;    }}void leftRotateString(char *s,int m,int n){    ReverseString(s, 0, m - 1);//需要反转的前半部分(比做成a数组),先自行反转    ReverseString(s, m, n-1);//后半部分(比做成b数组),在自行反转    ReverseString(s, 0, n - 1);//全体反转}int main(){    char s[100];    cin >> s;    leftRotateString(s, 3, strlen(s));    cout << s;    return 0;}//时间复杂度为O(N),空间为O(1)

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