| comments | difficulty | edit_url |
|---|---|---|
true |
简单 |
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
注意：本题与主站 21 题相同：https://leetcode.cn/problems/merge-two-sorted-lists/
我们先创建一个虚拟头结点 dummy，然后创建一个指针 cur 指向 dummy。
接下来，循环比较 l1 和 l2 的值，将较小的值接在 cur 后面，然后将指针向后移动一位。循环结束，将 cur 指向 l1 或者 l2 中剩余的部分。
最后返回 dummy.next 即可。
时间复杂度
# 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 = cur = ListNode(0)
while l1 and l2:
if l1.val <= l2.val:
cur.next = l1
l1 = l1.next
else:
cur.next = l2
l2 = l2.next
cur = cur.next
cur.next = l1 or l2
return dummy.next/**
* 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 cur = dummy;
while (l1 != null && l2 != null) {
if (l1.val <= l2.val) {
cur.next = l1;
l1 = l1.next;
} else {
cur.next = l2;
l2 = l2.next;
}
cur = cur.next;
}
cur.next = l1 == null ? l2 : l1;
return dummy.next;
}
}/**
* 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* dummy = new ListNode(0);
ListNode* cur = dummy;
while (l1 && l2) {
if (l1->val <= l2->val) {
cur->next = l1;
l1 = l1->next;
} else {
cur->next = l2;
l2 = l2->next;
}
cur = cur->next;
}
cur->next = l1 ? l1 : l2;
return dummy->next;
}
};/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
dummy := &ListNode{}
cur := dummy
for l1 != nil && l2 != nil {
if l1.Val <= l2.Val {
cur.Next = l1
l1 = l1.Next
} else {
cur.Next = l2
l2 = l2.Next
}
cur = cur.Next
}
if l1 == nil {
cur.Next = l2
} else {
cur.Next = l1
}
return dummy.Next
}/**
* Definition for singly-linked list.
* class ListNode {
* val: number
* next: ListNode | null
* constructor(val?: number, next?: ListNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
* }
*/
function mergeTwoLists(l1: ListNode | null, l2: ListNode | null): ListNode | null {
const dummy = new ListNode(0);
let cur = dummy;
while (l1 && l2) {
if (l1.val <= l2.val) {
cur.next = l1;
l1 = l1.next;
} else {
cur.next = l2;
l2 = l2.next;
}
cur = cur.next;
}
cur.next = l1 || l2;
return dummy.next;
}// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
// pub val: i32,
// pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
// #[inline]
// fn new(val: i32) -> Self {
// ListNode {
// next: None,
// val
// }
// }
// }
impl Solution {
pub fn merge_two_lists(
mut l1: Option<Box<ListNode>>,
mut l2: Option<Box<ListNode>>,
) -> Option<Box<ListNode>> {
match (l1.is_some(), l2.is_some()) {
(false, false) => None,
(true, false) => l1,
(false, true) => l2,
(true, true) => {
let mut dummy = Box::new(ListNode::new(0));
let mut cur = &mut dummy;
while l1.is_some() && l2.is_some() {
cur.next = if l1.as_ref().unwrap().val < l2.as_ref().unwrap().val {
let mut res = l1.take();
l1 = res.as_mut().unwrap().next.take();
res
} else {
let mut res = l2.take();
l2 = res.as_mut().unwrap().next.take();
res
};
cur = cur.next.as_mut().unwrap();
}
cur.next = if l1.is_some() { l1.take() } else { l2.take() };
dummy.next.take()
}
}
}
}/**
* 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) {
const dummy = new ListNode(0);
let cur = dummy;
while (l1 && l2) {
if (l1.val <= l2.val) {
cur.next = l1;
l1 = l1.next;
} else {
cur.next = l2;
l2 = l2.next;
}
cur = cur.next;
}
cur.next = l1 || l2;
return dummy.next;
};/**
* Definition for singly-linked list.
* public class ListNode {
* public int val;
* public ListNode next;
* public ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode MergeTwoLists(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(0);
ListNode cur = dummy;
while (l1 != null && l2 != null) {
if (l1.val <= l2.val) {
cur.next = l1;
l1 = l1.next;
} else {
cur.next = l2;
l2 = l2.next;
}
cur = cur.next;
}
cur.next = l1 == null ? l2 : l1;
return dummy.next;
}
}/* public class ListNode {
* var val: Int
* var next: ListNode?
* init(_ val: Int) {
* self.val = val
* self.next = nil
* }
* }
*/
class Solution {
func mergeTwoLists(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
let dummy = ListNode(0)
var cur: ListNode? = dummy
var l1 = l1
var l2 = l2
while let l1Node = l1, let l2Node = l2 {
if l1Node.val <= l2Node.val {
cur?.next = l1Node
l1 = l1Node.next
} else {
cur?.next = l2Node
l2 = l2Node.next
}
cur = cur?.next
}
cur?.next = l1 ?? l2
return dummy.next
}
}我们先判断 l1 和 l2 中有没有一个为空，如果有一个为空，那么我们直接返回另一个链表即可。
接下来，我们比较 l1 和 l2 的值：
- 如果
l1的值小于等于l2的值，我们递归调用mergeTwoLists(l1.next, l2)，并将l1.next指向返回的链表，然后返回l1。 - 如果
l1的值大于l2的值，我们递归调用mergeTwoLists(l1, l2.next)，并将l2.next指向返回的链表，然后返回l2。
时间复杂度
# 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 l1 is None or l2 is None:
return l1 or l2
if l1.val <= l2.val:
l1.next = self.mergeTwoLists(l1.next, l2)
return l1
else:
l2.next = self.mergeTwoLists(l1, l2.next)
return l2/**
* 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;
}
}
}/**
* 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) {
if (!l1) {
return l2;
}
if (!l2) {
return l1;
}
if (l1->val <= l2->val) {
l1->next = mergeTwoLists(l1->next, l2);
return l1;
} else {
l2->next = mergeTwoLists(l1, l2->next);
return l2;
}
}
};/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
if l1 == nil {
return l2
}
if l2 == nil {
return l1
}
if l1.Val <= l2.Val {
l1.Next = mergeTwoLists(l1.Next, l2)
return l1
} else {
l2.Next = mergeTwoLists(l1, l2.Next)
return l2
}
}/**
* Definition for singly-linked list.
* class ListNode {
* val: number
* next: ListNode | null
* constructor(val?: number, next?: ListNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
* }
*/
function mergeTwoLists(l1: ListNode | null, l2: ListNode | null): ListNode | null {
if (l1 == null || l2 == null) {
return l1 || l2;
}
if (l1.val < l2.val) {
l1.next = mergeTwoLists(l1.next, l2);
return l1;
}
l2.next = mergeTwoLists(l1, l2.next);
return l2;
}// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
// pub val: i32,
// pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
// #[inline]
// fn new(val: i32) -> Self {
// ListNode {
// next: None,
// val
// }
// }
// }
impl Solution {
pub fn merge_two_lists(
l1: Option<Box<ListNode>>,
l2: Option<Box<ListNode>>,
) -> Option<Box<ListNode>> {
match (l1, l2) {
(Some(mut n1), Some(mut n2)) => {
if n1.val < n2.val {
n1.next = Self::merge_two_lists(n1.next, Some(n2));
Some(n1)
} else {
n2.next = Self::merge_two_lists(Some(n1), n2.next);
Some(n2)
}
}
(Some(node), None) => Some(node),
(None, Some(node)) => Some(node),
(None, None) => None,
}
}
}/**
* 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 && l2)) {
return l1 || l2;
}
if (l1.val < l2.val) {
l1.next = mergeTwoLists(l1.next, l2);
return l1;
} else {
l2.next = mergeTwoLists(l2.next, l1);
return l2;
}
};/**
* Definition for singly-linked list.
* public class ListNode {
* public int val;
* public ListNode next;
* public ListNode(int x) { val = x; }
* }
*/
public 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;
}
}
}