剑指 Offer 25. 合并两个排序的链表
本问题对应的 leetcode 原文链接：剑指 Offer 25. 合并两个排序的链表
问题描述
输入两个递增排序的链表，合并这两个链表并使新链表中的节点仍然是递增排序的。
示例1：
输入：1->2->4, 1->3->4
输出：1->1->2->3->4->4
限制：
0 <= 链表长度 <= 1000
解题思路
视频讲解直达： 本题视频讲解
代码实现
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode merge = new ListNode(0);
ListNode temp = merge;
while(l1 != null && l2 != null){
if(l1.val < l2.val){
temp.next = l1;
l1 = l1.next;
} else {
temp.next = l2;
l2 = l2.next;
}
temp = temp.next;
}
temp.next = l1 == null ? l2 : l1;
return merge.next;
}
}
时间复杂度：O(n+m)，n 和 m 表示两个链表长度
空间复杂度：O(1)
Python
# 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
"""
merge = ListNode(0)
temp = merge
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 merge.next
C++
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
ListNode* merge = new ListNode(0);
ListNode* temp = merge;
while (l1 && 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 ? l1 : l2;
return merge->next;
}
};
Go
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
merge := &ListNode{}
temp := merge
for l1 != nil && l2 != nil {
if l1.Val < l2.Val {
temp.Next = l1
l1 = l1.Next
} else {
temp.Next = l2
l2 = l2.Next
}
temp = temp.Next
}
if l1 != nil {
temp.Next = l1
} else {
temp.Next = l2
}
return merge.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) {
let merge = new ListNode(0);
let temp = merge;
while (l1 != null && l2 != null) {
if (l1.val < l2.val) {
temp.next = l1;
l1 = l1.next;
} else {
temp.next = l2;
l2 = l2.next;
}
temp = temp.next;
}
temp.next = l1 == null ? l2 : l1;
return merge.next;
};