一、题目
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→…
二、思路
1）使用快慢指针找到中位点，将链表拆分成两个，然后倒转第二个链表，然后合并两个链表
2）使用双端队列
三、代码
1)
public void reorderList(ListNode head) {
ListNode res = head;
if (head == null || head.next == null) {
return;
}
ListNode slow = head;
ListNode fast = head;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
ListNode reverse = slow.next;
slow.next = null;
ListNode pre = null;
ListNode cur = reverse;
while (cur != null) {
ListNode next = cur.next;
cur.next = pre;
pre = cur;
cur = next;
}
reverse = pre;
ListNode cur1 = head;
ListNode cur2 = reverse;
while (cur1 != null && cur2 != null) {
ListNode next1 = cur1.next;
ListNode next2 = cur2.next;
cur2.next = next1;
cur1.next =cur2;
cur1 = next1;
cur2 = next2;
}
}
2)
public void reorderList(ListNode head) {
LinkedList<ListNode> queue = new LinkedList<>();
ListNode cur = head;
while (cur != null) {
queue.addLast(cur);
cur = cur.next;
}
while (!queue.isEmpty()) {
if (cur == null) {
cur = queue.pollFirst();
} else {
cur.next = queue.pollFirst();
cur = cur.next;
}
cur.next = queue.pollLast();
cur = cur.next;
}
if (cur != null) {
cur.next = null;
}
}
