Question

2130. 链表最大孪生和

English Version

题目描述

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

解法

方法一：链表转成列表（数组）求解

时间复杂度 $O(n)$，空间复杂度 $O(n)$。

# 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
  }
}

方法二：快慢指针 + 反转链表 + 双指针

时间复杂度 $O(n)$，空间复杂度 $O(1)$。

# 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;
}