C++ detalied solution with comments


ListNode* Solution::rotateRight(ListNode* A, int B) {
// If the list is empty or B is 0 return A.
if(A == NULL || B == 0){
return A;
ListNode *temp = A;
int count = 0;
// count no.of nodes in the given list
while(temp != NULL){
temp = temp->next;
// if given B is greater than no. of nodes in the list then, we find an effective number to shift the list by right.
int r = B%count;
// if r is 0 the list will end up with the same original head even after rotation.
if(r == 0 && B>=count){
return A;
int i = 0;
temp = A;
// Initialize a node to keep track of previous node.
ListNode *prev = NULL;
// Run a while loop to find the new_head node.
while(i<count - r){
prev = temp;
temp = temp->next;
ListNode *new_head = temp;
// After finding the new_head the prev node should be pointing to NULL since it is the tail node of the rotated list.
prev->next = NULL;
// run a while loop and to find the tail node of the original list.
temp = temp->next;
// Now the original tail node should be pointing to the original head node.
temp->next = A;
return new_head;