Merge Two Sorted Lists
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# 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

### 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.

## Solution

• 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.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 /** * 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; } };