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Problem Statement:
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.
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
list1andlist2are sorted in non-decreasing order.
- Time: O(|list1| + |list2|))O(∣list1∣+∣list2∣))
- Space: O(|list1| + |list2|))O(∣list1∣+∣list2∣))
Merge Two Sorted List Program Solution In C++
class Solution {
public:
ListNode* mergeTwoLists(ListNode* list1, ListNode* list2) {
if (!list1 || !list2)
return list1 ? list1 : list2;
if (list1->val > list2->val)
swap(list1, list2);
list1->next = mergeTwoLists(list1->next, list2);
return list1;
}
};
Merge Two Sorted List Program Solution In Java
class Solution {
public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
if (list1 == null || list2 == null)
return list1 == null ? list2 : list1;
if (list1.val > list2.val) {
ListNode temp = list1;
list1 = list2;
list2 = temp;
}
list1.next = mergeTwoLists(list1.next, list2);
return list1;
}
}
Merge Two Sorted List Program Solution In Python
class Solution:
def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
if not list1 or not list2:
return list1 if list1 else list2
if list1.val > list2.val:
list1, list2 = list2, list1
list1.next = self.mergeTwoLists(list1.next, list2)
return list1
