Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…
You may not modify the values in the list's nodes, only nodes itself may be changed.
Example 1:
Given 1->2->3->4, reorder it to 1->4->2->3.
Example 2:
Given 1->2->3->4->5, reorder it to 1->5->2->4->3.
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// OJ: https://leetcode.com/problems/reorder-list/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
private:
ListNode *reverse(ListNode* head) {
ListNode dummy;
while (head) {
auto node = head;
head = head->next;
node->next = dummy.next;
dummy.next = node;
}
return dummy.next;
}
int getLength(ListNode *head) {
int len = 0;
for (; head; head = head->next) ++len;
return len;
}
public:
void reorderList(ListNode* head) {
if (!head) return;
int len = (getLength(head) - 1) / 2;
ListNode *p = head, *q;
while (len--) p = p->next;
q = p->next;
p->next = NULL;
q = reverse(q);
p = head;
while (q) {
auto node = q;
q = q->next;
node->next = p->next;
p->next = node;
p = node->next;
}
}
};
// OJ: https://leetcode.com/problems/reorder-list/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
private:
ListNode *reverseList(ListNode *head) {
ListNode newHead;
while (head) {
auto p = head;
head = head->next;
p->next = newHead.next;
newHead.next = p;
}
return newHead.next;
}
public:
void reorderList(ListNode* head) {
if (!head) return;
ListNode *fast = head, *slow = head, *rightHead;
while (fast->next && fast->next->next) {
fast = fast->next->next;
slow = slow->next;
}
rightHead = slow->next;
slow->next = NULL;
rightHead = reverseList(rightHead);
while (rightHead) {
auto p = rightHead;
rightHead = rightHead->next;
p->next = head->next;
head->next = p;
head = p->next;
}
}
};