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Reverse Linked List II in C++
Suppose we have a linked list. We have to reverse the nodes from position m to n. We have to do it in one pass. So if the list is [1,2,3,4,5] and m = 2 and n = 4, then the result will be [1,4,,3,2,5]
Let us see the steps −
- There will be two methods, the reverseN() and reverseBetween(). The reverseBetween() will work as main method.
- define one link node pointer called successor as null
- The reverseN will work as follows −
- if n = 1, then successor := next of head, and return head
- last = reverseN(next of head, n - 1)
- next of (next of head) = head, and next of head := successor, return last
- the reverseBetween() method will be like −
- if m = 1, then return reverseN(head, n)
- next of head := reverseBetween(next of head, m – 1, n – 1)
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){
cout << ptr->val << ", ";
ptr = ptr->next;
}
cout << "]" << endl;
}
class Solution {
public:
ListNode* successor = NULL;
ListNode* reverseN(ListNode* head, int n ){
if(n == 1){
successor = head->next;
return head;
}
ListNode* last = reverseN(head->next, n - 1);
head->next->next = head;
head->next = successor;
return last;
}
ListNode* reverseBetween(ListNode* head, int m, int n) {
if(m == 1){
return reverseN(head, n);
}
head->next = reverseBetween(head->next, m - 1, n - 1);
return head;
}
};
main(){
Solution ob;
vector<int> v = {1,2,3,4,5,6,7,8};
ListNode *head = make_list(v);
print_list(ob.reverseBetween(head, 2, 6));
}Input
[1,2,3,4,5,6,7,8] 2 6
Output
[1, 6, 5, 4, 3, 2, 7, 8, ]
Updated on: 04-May-2020
160 Views
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