🌵21. Merge Two Sorted Lists
https://leetcode.com/problems/merge-two-sorted-lists/
js
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} l1
* @param {ListNode} l2
* @return {ListNode}
*/
var mergeTwoLists = function(l1, l2) {
var head = new ListNode()
var current = head
while (l1 && l2) {
if (l1.val <= l2.val) {
current.next = l1
l1 = l1.next
} else {
current.next = l2
l2 = l2.next
}
current = current.next
}
current.next = l1 ? l1 : l2
return head.next
}js
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} l1
* @param {ListNode} l2
* @return {ListNode}
*/
var mergeTwoLists = function(l1, l2) {
if (!l1) {
return l2
}
if (!l2) {
return l1
}
if (l1.val > l2.val) {
let temp = l1
l1 = l2
l2 = temp
}
l1.next = mergeTwoLists(l1.next, l2)
return l1
}py
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def mergeTwoLists(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
head = ListNode(None)
temp = head
while l1 and l2:
if l1.val <= l2.val:
temp.next = l1
l1 = l1.next
else:
temp.next = l2
l2 = l2.next
temp = temp.next
temp.next = l1 if l1 else l2
return head.nextpy
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def mergeTwoLists(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
if not l1:
return l2
if not l2:
return l1
if l1.val > l2.val:
l1, l2 = l2, l1
l1.next = self.mergeTwoLists(l1.next, l2)
return l1