字符串旋转

下面描述两种算法
空间复杂度都为O（1）

解法一

暴力移位

#define len(a) (sizeof(a)/sizeof(*a))
#include 
#include 
void leftShiftOne(char* string, int length);
void leftRotateString(char* string, int length, int num);
int main(void){
    char string[] = "asdfqwer";
    printf("%s\n", string);
    leftRotateString(string,len(string)-1,4);
    printf("%s\n", string);
}

void leftShiftOne(char* string, int length) {
    assert(string != NULL);
    char tmp = string[0];
    int i;
    for(i=1; i < length; i++) {
        string[i-1] = string[i];
    }
    string[length-1] = tmp;
}

void leftRotateString(char* string, int length, int num) {
    assert(num>=0);
    while(num--) {
        leftShiftOne(string,length);
    }
}

时间复杂度num*length
空间复杂度O(1)

解法二

三步反转法

它基于一个公式
X="abc"
X^T="cba"
(X^T Y^T)T=YX

#include 
#include 
#define len(a) (sizeof(a)/sizeof(*a))
void reverseString(char* s, int from ,int to);
void leftRotateString(char *s, int length, int num);
int main(void){
    char s[] = "asdfqwer";
    printf("%s\n",s);
    leftRotateString(s, len(s)-1, 3);
    printf("%s\n",s);
    return 0;
}

void reverseString(char* s, int from ,int to) {
    while(from < to) {
        char tmp = s[from];
        s[from++] = s[to];
        s[to--] = tmp;
    }
}

void leftRotateString(char *s, int length, int num) {
    num = num%length;
    reverseString(s,0,num-1);
    reverseString(s,num,length-1);
    reverseString(s,0,length-1);
}