给定一个单链表 L:L0→L1→…→Ln-1→Ln , 将其重新排列后变为: L0→Ln→L1→Ln-1→L2→Ln-2→…
你不能只是单纯的改变节点内部的值,而是需要实际的进行节点交换。
示例 1:
给定链表 1->2->3->4, 重新排列为 1->4->2->3.
示例 2:
给定链表 1->2->3->4->5, 重新排列为 1->5->2->4->3.
解题思路:
- 快慢指针找到中点,分为前半部分和后半部分两个链表
- 将后半部分的链表反转
- 将前半部分链表和后半部分链表间隔插入
Java代码:
class Solution {
public void reorderList(ListNode head) {
ListNode slow = head, fast = head, p, q, tmp = new ListNode(0x3FFFFF);
tmp.next = head;
while (null != fast && null != fast.next) {
tmp = tmp.next;
slow = slow.next;
fast = fast.next.next;
}
tmp.next = null;
fast = slow;
slow = null;
while (null != fast) {
p = slow;
slow = fast;
fast = fast.next;
slow.next = p;
}
fast = head;
while (null != slow && null != fast) {
p = fast.next;
q = slow.next;
fast.next = slow;
slow.next = p;
tmp = slow;
fast = p;
slow = q;
}
tmp.next = slow;
}
}