【LetMeFly】21.合并两个有序链表
力扣题目链接:https://leetcode.cn/problems/merge-two-sorted-lists/
将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。
示例 1:
输入:l1 = [1,2,4], l2 = [1,3,4] 输出:[1,1,2,3,4,4]
示例 2:
输入:l1 = [], l2 = [] 输出:[]
示例 3:
输入:l1 = [], l2 = [0] 输出:[0]
提示:
- 两个链表的节点数目范围是
[0, 50]
-100 <= Node.val <= 100
l1
和l2
均按 非递减顺序 排列
方法一:遍历
使用一个指针p指向新的链表,
在list1和list2都非空时:
-
若 l i s t 1 − > v a l < l i s t 2 − > v a l list1->val < list2->val list1−>val<list2−>val,则 p − > n e x t = l i s t 1 , l i s t 1 = l i s t 1 − > n e x t p->next = list1, list1 = list1->next p−>next=list1,list1=list1−>next
-
否则, p − > n e x t = l i s t 2 , l i s t 2 = l i s t 2 − > n e x t p->next = list2, list2 = list2->next p−>next=list2,list2=list2−>next
-
无论如何都: p = p − > n e x t p = p->next p=p−>next
-
时间复杂度 O ( l e n ( l i s t 1 ) + l e n ( l i s t 2 ) ) O(len(list1) + len(list2)) O(len(list1)+len(list2))(其实为 min ( l e n ( l i s t 1 ) , l e n ( l i s t 2 ) ) \min(len(list1), len(list2)) min(len(list1),len(list2))
-
空间复杂度 O ( 1 ) O(1) O(1),力扣返回值不计入算法空间复杂度。(如果list1和list2允许修改的话,空间使用量将会更低)
AC代码
C++
class Solution {
public:
ListNode* mergeTwoLists(ListNode* list1, ListNode* list2) {
ListNode* head = new ListNode();
ListNode* p = head;
while (list1 && list2) {
if (list1->val < list2->val) {
p->next = list1;
list1 = list1->next;
}
else {
p->next = list2;
// list2->=next; (笑)
list2 = list2->next;
}
p = p->next;
}
if (list1) {
p->next = list1;
}
else {
p->next = list2;
}
return head->next;
}
};
Python
# from typing import Optional
# # Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
head = ListNode()
p = head
while list1 and list2:
if list1.val < list2.val:
p.next = list1
list1 = list1.next
else:
p.next = list2
list2 = list2.next
p = p.next
if list1:
p.next = list1
else:
p.next = list2
return head.next
同步发文于CSDN,原创不易,转载请附上原文链接哦~
Tisfy:https://letmefly.blog.csdn.net/article/details/132115272