Sort List in C++

C++Server Side ProgrammingProgramming

Suppose we have a list, we have to sort this in O(n logn) time using constant space complexity, so if the list is like [4,2,1,3], then it will be [1,2,3,4]

To solve this, we will follow these steps −

  • Define a method for merging two lists in sorted, order, that method is merge(), this takes two lists l1 and l2.

  • The sort list method will work like below −

  • if head is null, or next of head is null, then return head

  • slow := head, fast := head, and prev = null

  • while fast is not null and next of fast is also not null, do

    • prev := slow

    • slow := next of slow

    • fast := next of next of fast

  • next of prev := null

  • l1 := sortList(head)

  • l2 := sortList(slow)

  • return merge(l1, l2)

Let us see the following implementation to get better understanding −

Example

 Live Demo

#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* sortList(ListNode* head) {
      if(!head || !head->next)return head;
      ListNode *slow = head, *fast = head, *prev = NULL;
      while(fast && fast->next){
         prev = slow;
         slow = slow->next;
         fast = fast->next->next;
      }
      prev->next = NULL;
      ListNode* l1 = sortList(head);
      ListNode* l2 = sortList(slow);
      return mergeList(l1,l2);
   }
   ListNode* mergeList(ListNode* l1, ListNode* l2){
      ListNode* temp = new ListNode(0);
      ListNode* p =temp;
      while(l1 && l2){
         if(l1->val<=l2->val){
            p->next = l1;
            l1 = l1->next;
         }else{
            p->next = l2;
            l2 = l2->next;
         }
         p = p->next;
      }
      if(l1){
         p->next = l1;
      }
      if(l2){
         p->next = l2;
      }
      return temp->next;
   }
};
main(){
   vector<int> v = {4,2,1,3,5,19,18,6,7};
   ListNode *h1 = make_list(v);
   Solution ob;
   print_list((ob.sortList(h1)));
}

Input

[4,2,1,3,5,9,8,6,7]

Output

[1, 2, 3, 4, 5, 6, 7, 18, 19, ]
raja
Published on 04-Mar-2020 05:56:33
Advertisements