LeetCode reorder-list

题目描述

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}.

解题思路：

• 链表拆分成左右两段
• 右半边逆序
• 依次拼接 左边走一个  右边走一个

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public void reorderList(ListNode head) {
        if (head == null) 
            return;
        ListNode[] ls = cut(head);
		ListNode left = ls[0];
		ListNode right = reverseList(ls[1]);
		ListNode preRightNext = null, preLeftNext = null, tmpLeft = left, tmpRight = right;
		while (tmpLeft != null || tmpRight != null) {
			if (tmpRight == null) {
				break;
			} else {
				preRightNext = tmpRight.next;
				preLeftNext = tmpLeft.next;
				tmpLeft.next = tmpRight;
				tmpRight.next = preLeftNext;
				tmpRight = preRightNext;
				tmpLeft = preLeftNext;
			}
		}
    }
    
    public ListNode[] cut(ListNode head) {
		ListNode fast = head, slow = head;
		while (fast.next != null && fast.next.next != null) {
			fast = fast.next.next;
			slow = slow.next;
		}
		ListNode right = slow.next;
		slow.next = null;
		return new ListNode[] {head, right};
	}
    
    public ListNode reverseList(ListNode head) {
		ListNode next = null;
		ListNode pNext = null;
		while (head != null) {
			pNext = head.next;
			head.next = next;
			next = head;
			head = pNext;
		}
		return next;
	}
}