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 node0
is the twin of node3
, and node1
is 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
Code
Time Complexity: , Space Complexity:
和 Reorder List 類似,先切兩半,反轉後半段。
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
int pairSum(struct ListNode* head) {
struct ListNode* slow = head;
struct ListNode* fast = head->next;
while(fast && fast->next) {
slow = slow->next;
fast = fast->next->next;
}
// isolate the frist half
struct ListNode* temp = slow->next;
slow->next = NULL;
// reverse the second half
struct ListNode* prev = NULL;
while(temp) {
struct ListNode* next = temp->next;
temp->next = prev;
prev = temp;
temp = next;
}
int res = INT_MIN;
while(prev && head) {
int sum = prev->val + head->val;
if(sum > res) {
res = sum;
}
prev = prev->next;
head = head->next;
}
return res;
}