在一个大小为 n 且 n 为 偶数 的链表中，对于 0 <= i <= (n / 2) - 1 的 i ，第 i 个节点（下标从 0 开始）的孪生节点为第 (n-1-i) 个节点 。
- 比方说，
n = 4那么节点0是节点3的孪生节点，节点1是节点2的孪生节点。这是长度为n = 4的链表中所有的孪生节点。
孪生和 定义为一个节点和它孪生节点两者值之和。
给你一个长度为偶数的链表的头节点 head ，请你返回链表的 最大孪生和 。
示例 1：
输入：head = [5,4,2,1] 输出：6 解释： 节点 0 和节点 1 分别是节点 3 和 2 的孪生节点。孪生和都为 6 。 链表中没有其他孪生节点。 所以，链表的最大孪生和是 6 。
示例 2：
输入：head = [4,2,2,3] 输出：7 解释： 链表中的孪生节点为： - 节点 0 是节点 3 的孪生节点，孪生和为 4 + 3 = 7 。 - 节点 1 是节点 2 的孪生节点，孪生和为 2 + 2 = 4 。 所以，最大孪生和为 max(7, 4) = 7 。
示例 3：
输入：head = [1,100000] 输出：100001 解释： 链表中只有一对孪生节点，孪生和为 1 + 100000 = 100001 。
提示：
- 链表的节点数目是
[2, 105]中的 偶数 。 1 <= Node.val <= 105
时间复杂度
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def pairSum(self, head: Optional[ListNode]) -> int:
s = []
while head:
s.append(head.val)
head = head.next
n = len(s)
return max(s[i] + s[-(i + 1)] for i in range(n >> 1))/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public int pairSum(ListNode head) {
List<Integer> s = new ArrayList<>();
for (; head != null; head = head.next) {
s.add(head.val);
}
int ans = 0, n = s.size();
for (int i = 0; i < (n >> 1); ++i) {
ans = Math.max(ans, s.get(i) + s.get(n - 1 - i));
}
return ans;
}
}/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
int pairSum(ListNode* head) {
vector<int> s;
for (; head != nullptr; head = head->next) s.push_back(head->val);
int ans = 0, n = s.size();
for (int i = 0; i < (n >> 1); ++i) ans = max(ans, s[i] + s[n - i - 1]);
return ans;
}
};/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func pairSum(head *ListNode) int {
var s []int
for ; head != nil; head = head.Next {
s = append(s, head.Val)
}
ans, n := 0, len(s)
for i := 0; i < (n >> 1); i++ {
if ans < s[i]+s[n-i-1] {
ans = s[i] + s[n-i-1]
}
}
return ans
}/**
* 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 pairSum(head: ListNode | null): number {
const arr = [];
let node = head;
while (node) {
arr.push(node.val);
node = node.next;
}
const n = arr.length;
let ans = 0;
for (let i = 0; i < n >> 1; i++) {
ans = Math.max(ans, arr[i] + arr[n - 1 - i]);
}
return ans;
}// 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 pair_sum(head: Option<Box<ListNode>>) -> i32 {
let mut arr = Vec::new();
let mut node = &head;
while node.is_some() {
let t = node.as_ref().unwrap();
arr.push(t.val);
node = &t.next;
}
let n = arr.len();
let mut ans = 0;
for i in 0..n >> 1 {
ans = ans.max(arr[i] + arr[n - 1 - i]);
}
ans
}
}时间复杂度
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def pairSum(self, head: Optional[ListNode]) -> int:
def reverse(head):
dummy = ListNode()
curr = head
while curr:
next = curr.next
curr.next = dummy.next
dummy.next = curr
curr = next
return dummy.next
slow, fast = head, head.next
while fast and fast.next:
slow, fast = slow.next, fast.next.next
pa = head
q = slow.next
slow.next = None
pb = reverse(q)
ans = 0
while pa and pb:
ans = max(ans, pa.val + pb.val)
pa = pa.next
pb = pb.next
return ans/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public int pairSum(ListNode head) {
ListNode slow = head;
ListNode fast = head.next;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
ListNode pa = head;
ListNode q = slow.next;
slow.next = null;
ListNode pb = reverse(q);
int ans = 0;
while (pa != null) {
ans = Math.max(ans, pa.val + pb.val);
pa = pa.next;
pb = pb.next;
}
return ans;
}
private ListNode reverse(ListNode head) {
ListNode dummy = new ListNode();
ListNode curr = head;
while (curr != null) {
ListNode next = curr.next;
curr.next = dummy.next;
dummy.next = curr;
curr = next;
}
return dummy.next;
}
}/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
int pairSum(ListNode* head) {
ListNode* slow = head;
ListNode* fast = head->next;
while (fast && fast->next) {
slow = slow->next;
fast = fast->next->next;
}
ListNode* pa = head;
ListNode* q = slow->next;
slow->next = nullptr;
ListNode* pb = reverse(q);
int ans = 0;
while (pa) {
ans = max(ans, pa->val + pb->val);
pa = pa->next;
pb = pb->next;
}
return ans;
}
ListNode* reverse(ListNode* head) {
ListNode* dummy = new ListNode();
ListNode* curr = head;
while (curr) {
ListNode* next = curr->next;
curr->next = dummy->next;
dummy->next = curr;
curr = next;
}
return dummy->next;
}
};/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func pairSum(head *ListNode) int {
reverse := func(head *ListNode) *ListNode {
dummy := &ListNode{}
curr := head
for curr != nil {
next := curr.Next
curr.Next = dummy.Next
dummy.Next = curr
curr = next
}
return dummy.Next
}
slow, fast := head, head.Next
for fast != nil && fast.Next != nil {
slow, fast = slow.Next, fast.Next.Next
}
pa := head
q := slow.Next
slow.Next = nil
pb := reverse(q)
ans := 0
for pa != nil {
ans = max(ans, pa.Val+pb.Val)
pa = pa.Next
pb = pb.Next
}
return ans
}/**
* 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 pairSum(head: ListNode | null): number {
let fast = head;
let slow = head;
while (fast) {
fast = fast.next.next;
slow = slow.next;
}
let prev = null;
while (slow) {
const next = slow.next;
slow.next = prev;
prev = slow;
slow = next;
}
let left = head;
let right = prev;
let ans = 0;
while (left && right) {
ans = Math.max(ans, left.val + right.val);
left = left.next;
right = right.next;
}
return ans;
}