[LeetCode] 21. 合并两个有序链表
题目链接:https://leetcode-cn.com/problems/merge-two-sorted-lists/
题目描述:
将两个有序链表合并为一个新的有序链表并返回。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:
输入:1->2->4, 1->3->4
输出:1->1->2->3->4->4
思路:
思路1:
迭代方法.
每次选两个链表头结点最小的,比如:我们生活中,有两个已经按照高矮排好的队伍,我们如何把变成一个队伍!当然,每次选两个队伍排头的,比较他们的高矮!组成新的的队伍.
时间复杂度:\(O(m+n)\)
空间复杂度:\(O(m+n)\)
思路2:
递归方法
关注我的知乎专栏,了解更多的解题技巧!
代码:
python
迭代
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
dummy = ListNode(0)
p = dummy
while l1 or l2:
if l1 and l2:
tmp1 = l1.val
tmp2 = l2.val
if tmp1 < tmp2:
p.next = ListNode(tmp1)
l1 = l1.next
else:
p.next = ListNode(tmp2)
l2 = l2.next
elif l1:
p.next = ListNode(l1.val)
l1 = l1.next
elif l2:
p.next = ListNode(l2.val)
l2 = l2.next
p = p.next
return dummy.next
java
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(0);
ListNode p = dummy;
while (l1 != null && l2 != null){
if (l1.val < l2.val){
p.next = l1;
l1 = l1.next;
}
else {
p.next = l2;
l2 = l2.next;
}
p = p.next;
}
if (l1 == null) p.next = l2;
if (l2 == null) p.next = l1;
return dummy.next;
}
}
递归
python
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
if not l1: return l2
if not l2: return l1
if l1.val < l2.val:
l1.next = self.mergeTwoLists(l1.next,l2)
return l1
else:
l2.next = self.mergeTwoLists(l1,l2.next)
return l2
java
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1 == null) return l2;
if (l2 == null) return l1;
if (l1.val < l2.val) {
l1.next = mergeTwoLists(l1.next,l2);
return l1;
}
else{
l2.next = mergeTwoLists(l1,l2.next);
return l2;
}
}