题目:
Given a singly linked list L: L 0→L 1→…→L n-1→L n,
reorder it to: L 0→L n →L 1→L n-1→L 2→L n-2→…
You must do this in-place without altering the nodes' values.
For example,
Given{1,2,3,4}, reorder it to{1,4,2,3}.
给定单链表L:L0→L 1→…→L_n-1→L_n,
将其重新排序为:l_0→L_n→L_1→L_n-1→L_2→L_n-2→…
你必须在不改变节点值的情况下就地执行此操作。
例如,给定{1,2,3,4},将其重新排序为{1,4,2,3}
思路:
分三部分
- 找出中点,拆分列表
- 中点后的链表反转
- 按要求合并链表
public class Solution {
public void reorderList(ListNode head) {
if(head == null || head.next == null)
return;
// 找中点
ListNode mid = findMiddle(head);
// 中点后的链表反转
ListNode rev = reverse(mid);
// 两个链表合并
merge(head,rev);
}
private static ListNode findMiddle(ListNode head){
ListNode slow = head;
ListNode fast = head;
while(fast.next != null && fast.next.next != null){
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
private static ListNode reverse(ListNode cur){
ListNode pre = null;
ListNode next = cur;
while(next != null){
next = cur.next;
cur.next = pre;
pre = cur;
cur = next;
}
return pre;
}
private static void merge(ListNode head, ListNode rev){
ListNode headNext = null;
ListNode revNext = null;
while(head != null && rev != null){
headNext = head.next;
revNext = rev.next;
head.next = rev;
rev.next = headNext;
head = headNext;
rev = revNext;
}
}
}