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_143_ReorderList.java
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_143_ReorderList.java
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package leetcode_1To300;
/**
* 本代码来自 Cspiration,由 @Cspiration 提供
* 题目来源:http://leetcode.com
* - Cspiration 致力于在 CS 领域内帮助中国人找到工作,让更多海外国人受益
* - 现有课程:Leetcode Java 版本视频讲解(1-900题)(上)(中)(下)三部
* - 算法基础知识(上)(下)两部;题型技巧讲解(上)(下)两部
* - 节省刷题时间,效率提高2-3倍,初学者轻松一天10题,入门者轻松一天20题
* - 讲师:Edward Shi
* - 官方网站:https://cspiration.com
* - 版权所有,转发请注明出处
*/
public class _143_ReorderList {
/**
* 143. Reorder List
* Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-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}.
time : O(n)
space : O(1)
* @param head
*/
public void reorderList(ListNode head) {
if (head == null || head.next == null) return;
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode temp = null;
ListNode slow = head, fast = head;
ListNode l1 = head;
while (fast != null && fast.next != null) {
temp = slow;
slow = slow.next;
fast = fast.next.next;
}
temp.next = null;
ListNode l2 = reverse(slow);
merge(l1, l2);
}
private ListNode reverse(ListNode head) {
ListNode prev = null;
while (head != null) {
ListNode temp = head.next;
head.next = prev;
prev = head;
head = temp;
}
return prev;
}
private void merge(ListNode l1, ListNode l2) {
while (l1 != l2) {
ListNode n1 = l1.next;
ListNode n2 = l2.next;
l1.next = l2;
if (n1 == null) break;
l2.next = n1;
l1 = n1;
l2 = n2;
}
}
}