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Program to rotate a linked list by k places in C++
Suppose we have a linked list. We have to rotate the list to the right by k places. The value of k will be positive. So if the list is like [1 −> 2 −> 3 −> 4 −> 5 −> NULL], and k = 2, then the output will be [4 −> 5 −> 1 −> 2 −> 3 −> NULL]
Let us see the steps −
If the list is empty, then return null
len := 1
create one node called tail := head
while next of tail is not null
increase len by 1
tail := next of tail
next of tail := head
k := k mod len
newHead := null
for i := 0 to len − k
tail := next of tail
newHead := next of tail
next of tail := null
return newHead
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
class ListNode{
public:
int val;
ListNode *next;
ListNode(int data){
val = data;
next = NULL;
}
};
ListNode *make_list(vector<int> v){
ListNode *head = new ListNode(v[0]);
for(int i = 1; i<v.size(); i++){
ListNode *ptr = head;
while(ptr->next != NULL){
ptr = ptr->next;
}
ptr->next = new ListNode(v[i]);
}
return head;
}
void print_list(ListNode *head){
ListNode *ptr = head;
cout << "[";
while(ptr->next){
cout << ptr->val << ", ";
ptr = ptr->next;
}
cout << "]" << endl;
}
class Solution {
public:
ListNode* rotateRight(ListNode* head, int k) {
if(!head) return head;
int len = 1;
ListNode* tail = head;
while(tail->next){
len++;
tail = tail->next;
}
tail->next = head;
k %= len;
ListNode* newHead = NULL;
for(int i = 0; i < len - k; i++){
tail = tail->next;
}
newHead = tail->next;
tail->next = NULL;
return newHead;
}
};
main(){
Solution ob;
vector<int> v = {1,2,3,4,5,6,7,8,9};
ListNode *head = make_list(v);
print_list(ob.rotateRight(head, 4));
}Input
[1,2,3,4,5,6,7,8,9], 4
Output
[6, 7, 8, 9, 1, 2, 3, 4, ]
Updated on: 21-Oct-2020
216 Views
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