Featured image of post LeetCode 75 - 2130. Maximum Twin Sum of a Linked List

LeetCode 75 - 2130. Maximum Twin Sum of a Linked List

Description

In a linked list of size n, where n is even, the ith node (0-indexed) of the linked list is known as the twin of the (n-1-i)th node, if 0 <= i <= (n / 2) - 1.

  • For example, if n = 4, then node 0 is the twin of node 3, and node 1 is the twin of node 2. These are the only nodes with twins for n = 4.

The twin sum is defined as the sum of a node and its twin.

Given the head of a linked list with even length, return the maximum twin sum of the linked list.

Example 1:

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Input: head = [5,4,2,1]
Output: 6
Explanation:
Nodes 0 and 1 are the twins of nodes 3 and 2, respectively. All have twin sum = 6.
There are no other nodes with twins in the linked list.
Thus, the maximum twin sum of the linked list is 6. 

Example 2:

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Input: head = [4,2,2,3]
Output: 7
Explanation:
The nodes with twins present in this linked list are:
- Node 0 is the twin of node 3 having a twin sum of 4 + 3 = 7.
- Node 1 is the twin of node 2 having a twin sum of 2 + 2 = 4.
Thus, the maximum twin sum of the linked list is max(7, 4) = 7. 

Example 3:

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Input: head = [1,100000]
Output: 100001
Explanation:
There is only one node with a twin in the linked list having twin sum of 1 + 100000 = 100001.

Constraints:

  • The number of nodes in the list is an even integer in the range [2, 105].
  • 1 <= Node.val <= 105

Solutions

Solution 1

Use the array directly to store the data for the corresponding node, and then look for the maximum twin sum.

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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) {
        int sum{};
        ListNode* current = head;
        std::vector<int> vec;

        while (current) {
            vec.push_back(current->val);
            current = current->next;
        }
        int n = vec.size();
        for (int i = 0; i < n; i++) {
            auto re = vec[i] + vec[n - 1 - i];

            sum = std::max(sum, re);
        }
        return sum;
    }
};

Solution 2

Fast/slow pointer+reverse linked list

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class Solution {
public:
    int pairSum(ListNode* head) {
        ListNode* slow = head;
        ListNode* fast = head->next;
        while (fast->next) {
            slow = slow->next;
            fast = fast->next->next;
        }
        // 反转链表
        ListNode* last = slow->next;
        while (last->next) {
            ListNode* cur = last->next;
            last->next = cur->next;
            cur->next = slow->next;
            slow->next = cur;
        }
        int ans = 0;
        ListNode* x = head;
        ListNode* y = slow->next;
        while (y) {
            ans = max(ans, x->val + y->val);
            x = x->next;
            y = y->next;
        }
        return ans;
    }
};