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2130. 链表最大孪生和

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：模拟

我们可以将链表中的节点值依次存入数组中，然后使用双指针分别指向数组的开头和结尾，计算每对孪生节点的孪生和，取最大值即为答案。

时间复杂度 \(O(n)\)，空间复杂度 \(O(n)\)。其中 \(n\) 是链表的节点数。

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# 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))

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/**
 * 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;
    }
}

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/**
 * 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;
    }
};

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/**
 * 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
}

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/**
 * 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;
}

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// 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 \(O(1)\)。

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# 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

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/**
 * 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;
    }
}

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/**
 * 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;
    }
};

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/**
 * 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
}

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/**
 * 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;
}

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