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 node0is the twin of node3, and node1is the twin of node2. These are the only nodes with twins forn = 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:
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:
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:
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
Solution: Two Pointers + Reverse List
Use fast slow pointers to find the middle point and reverse the second half.
Time complexity: O(n)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: int pairSum(ListNode* head) { auto reverse = [](ListNode* head, ListNode* prev = nullptr) { while (head) { swap(head->next, prev); swap(head, prev); } return prev; }; ListNode* fast = head; ListNode* slow = head; while (fast) { fast = fast->next->next; slow = slow->next; } slow = reverse(slow); int ans = 0; while (slow) { ans = max(ans, head->val + slow->val); head = head->next; slow = slow->next; } return ans; } }; |
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