**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.

- We compare the values at the pointer position and we point the
- 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.

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/**
* 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;
}
};