Rotate K strings to the left (details)

         Today I will bring you a very interesting question Rotate to the left k A string , The conventional solution is very simple , But what I'm talking about today is another interesting method , It may be a great inspiration for you to brush questions in the future .

One . Violence solution :

        Violence solution : first floor for loop , The number of overall traversal is to rotate several strings , The second floor for The cycle is the number of exchanges .

#include<stdio.h>
#include<string.h>
void left_rotate(char arr[], int k) {
	int len = strlen(arr);
	k %= len;// No matter what k How big , Search times are inevitable <len-1
	for (int i = 0; i < k; i++) {
		char tmp = arr[0];
		for (int j = 0; j < len-1; j++) {
			arr[j] = arr[j+1];
		}
		arr[len-1] = tmp;
	}
}
int main() {
	char arr[] = "abcdef";
	int k = 0;
	scanf_s("%d", &k);
	left_rotate(arr, k);
	printf("%s", arr);
	return 0;
}

Two . Novel solution :

         Suppose we have a string :"abcdef", And rotate two strings to the left . So we can put "abcdef" It is divided into "ab","cdef" First rotate separately "ba" and "fedc" Turn into "bafedc", Then the whole rotation becomes "cdefab". Isn't it amazing ?

 

#include<stdio.h>
#include<string.h>
#include<assert.h>
// Method 2 :
void reverse(char* left, char* right) {
	assert(left && right);
	while (left < right) {
		char tmp = *left;
		*left = *right;
		*right = tmp;
		left++;
		right--;
	}
}
void left_rotate(char arr[],int k,size_t len) {
	k %= len;
	reverse(arr, arr + k - 1);
	reverse(arr + k, arr + len - 1);
	reverse(arr, arr + len - 1);
}
int main() {
	char arr[] = "abcdef";
	int k = 0;
	size_t len = strlen(arr);
	scanf_s("%d", &k);
	left_rotate(arr,k,len);
	printf("%s", arr);
	return 0;
}

summary :

        The above are the two solutions I brought. I hope they will be helpful to your follow-up study .