Leetcode Merge-Two-Sorted-Lists

2022-06-14 22:57

Algorithms:

#algorithm #two_pointers

Data structures:

#DS #linked_list

Difficulty:

#leetcode #coding_problem #difficulty-easy

tag:#two_pointers

Problem

You are given the heads of two sorted linked lists list1 and list2.

Merge the two lists in a one sorted list. The list should be made by splicing together the nodes of the first two lists.

Return the head of the merged linked list.

Examples

Example 1:

**Input:** list1 = [1,2,4], list2 = [1,3,4]
**Output:** [1,1,2,3,4,4]

Example 2:

**Input:** list1 = [], list2 = []
**Output:** []

Example 3:

**Input:** list1 = [], list2 = [0]
**Output:** [0]

Constraints

The number of nodes in both lists is in the range [0, 50].
-100 <= Node.val <= 100
Both list1 and list2 are sorted in non-decreasing order.

Thoughts

This is a #two_pointers algorithm, I've done similar problems at leetcode's array list. The only thing to watch out for is when there is one list remaining, remember to add the tails.

Solution

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
  ListNode *mergeTwoLists(ListNode *list1, ListNode *list2) {
    // 2 Pointers, Space and time O(m + n);
    ListNode *ptr1 = list1;
    ListNode *ptr2 = list2;
    ListNode *dummyHead = new ListNode();
    ListNode *ptr3 = dummyHead;
    while (ptr2 != NULL && ptr1 != NULL) {

      if (ptr2->val <= ptr1->val) {
        ptr3->next = ptr2;
        ptr2 = ptr2->next;
      } else {
        ptr3->next = ptr1;
        ptr1 = ptr1->next;
      }

      ptr3 = ptr3->next;
    }

    if (ptr2 == NULL) {
      ptr3->next = ptr1;
    } else if (ptr1 == NULL) {
      ptr3->next = ptr2;
    }

    return dummyHead->next;
  }
};

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