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| class Solution {
public void reorderList(ListNode head) {
if (head == null || head.next == null) {
return;
}
ListNode slow = head, fast = head;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
ListNode left = head, right = reverseList(slow), dummy = new ListNode(0), ptr = dummy;
while (left != right && right != null) {
ptr.next = left;
left = left.next;
ptr = ptr.next;
ptr.next = right;
right = right.next;
ptr = ptr.next;
}
if (left == right) {
ptr.next = left;
}
}
private ListNode reverseList(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode prev = null, curr = head;
while (curr != null) {
ListNode next = curr.next;
curr.next = prev;
prev = curr;
curr = next;
}
return prev;
}
}
|
快慢指针, 再reverse list再merge.
时间复杂度: O(n)
空间复杂度: O(1)
