LeetCode #21: Merge Two Sorted Lists (C/C++).

easy

source: https://leetcode.com/problems/merge-two-sorted-lists/
C/C++ Solution to LeetCode problem 21. Merge Two Sorted Lists.

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

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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

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• 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.

Solution

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• We will use the list with the initial smaller node value as our head.
• We will have a tmp pointer to keep traking of the element that is next to the one we will sort.

• We will move two pointers, one for each list.

  • We compare the values at the pointer position and we point the ->nexr value to the node with the bigger value.
  • The bigger value now will point to the value that was saved in our tmp pointer (the one that our smaller value was pointing before in ->next).
  • We move the pointer of the list with the bigger value.
• Once one of the pointers reached the end, we move to the other list and just keep adding the node to the end of the already sorted list.

/**
 * 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) {
    if (!list1 || !list2)
      return list1 ? list1: list2;

    ListNode* mergedHead = nullptr;
    ListNode* p0 = nullptr;
    ListNode* p1 = list1;
    ListNode* p2 = list2;
    ListNode *tmp;

    if (p1->val < p2->val) {
      mergedHead = list1;
      p1 = p1->next;
    } else {
      mergedHead = list2;
      p2 = p2->next;
    }
    p0 = mergedHead;

    while(p1 && p2) {
      if (p1->val < p2->val) {
        tmp = p1;
        p1 = p1->next;
      } else {
        tmp = p2;
        p2 = p2->next;
      }

      p0->next = tmp;
      p0 = p0->next;
    }

    tmp = p1 ? p1 : p2;
    while(tmp) {
      p0->next = tmp;
      tmp = tmp->next;
      p0 = p0->next;
    }
    
    return mergedHead;
  }
};