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/
同时遍历两个链表,归并插入新链表中即可。
# 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 and l2:
if l1.val <= l2.val:
p.next = l1
l1 = l1.next
else:
p.next = l2
l2 = l2.next
p = p.next
p.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 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;
}
p.next = l1 == null ? l2 : l1;
return dummy.next;
}
}
- 递归
/**
* 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;
}
};
- 遍历
/**
* 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 res = new ListNode();
let cur = res;
while (l1 && l2) {
let node;
if (l1.val < l2.val) {
node = l1;
l1 = l1.next;
} else {
node = l2;
l2 = l2.next;
}
cur.next = node;
cur = node;
}
cur.next = l1 || l2;
return res.next;
};
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
}
l2.Next = mergeTwoLists(l1, l2.Next)
return l2
}
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
if (nullptr == l1 && nullptr == l2) {
return nullptr; // 两个都为空,则直接返回
}
if (nullptr == l1 || nullptr == l2) {
return l1 == nullptr ? l2 : l1; // 有且仅有一个为空,则返回非空节点
}
ListNode* node = nullptr;
if (l1->val > l2->val) {
node = l2;
node->next = mergeTwoLists(l1, l2->next);
} else {
node = l1;
node->next = mergeTwoLists(l1->next, l2);
}
return node;
}
};
/**
* 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 duumy = new ListNode();
let cur = duumy;
while (l1 && l2) {
let node: ListNode;
if (l1.val < l2.val) {
node = l1;
l1 = l1.next;
} else {
node = l2;
l2 = l2.next;
}
cur.next = node;
cur = node;
}
cur.next = l1 || l2;
return duumy.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 {
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.
// #[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()
}
}
}
}