【LeetCode】 143. Reorder List 重排链表（Medium）（JAVA）
题目地址: https://leetcode.com/problems/reorder-list/
题目描述：
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.
题目大意
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。
解题方法
1、快慢指针找出中间节点
2、中间节点之后的所有元素翻转
3、从头开始遍历一个个元素接起来
class Solution {
public void reorderList(ListNode head) {
if (head == null || head.next == null) return;
ListNode fast = head.next;
ListNode slow = head;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
ListNode start = slow.next;
slow.next = null;
ListNode reverse = null;
while (start != null) {
ListNode next = start.next;
start.next = reverse;
reverse = start;
start = next;
}
start = head;
while (start != null && reverse != null) {
ListNode next = start.next;
start.next = reverse;
reverse = reverse.next;
start.next.next = next;
start = next;
}
}
}
执行耗时:1 ms,击败了100.00% 的Java用户
内存消耗:40.5 MB,击败了99.74% 的Java用户