reorder-list-LeetCode

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 must do this in-place without altering the nodes' values. For example, Given{1,2,3,4}, reorder it to{1,4,2,3}.

思路

1. 根据题意，一个链表是从后往前相间的从前往后插入链表，可以将链表分为两部分：前面的从先往后，后面一部分从后往前，这样，后面一部分就符合“栈”的存储结构
2. 想用先后两个步数不同的指针将链表分为两部分，这样分出来的结果根据链表节点的奇偶不同，后面链表的个数小于等于前面的。然后用栈存储后面的链表，再遍历前面的链表将节点插入。

代码

import java.util.ArrayDeque;
/**
 * 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 || head.next == null) {
			return;
		}
		ArrayDeque<ListNode> stack = new ArrayDeque<ListNode>();
		ListNode pre = head;
		ListNode follow = head.next;
		while (follow.next != null) {
			pre = pre.next;
			follow = follow.next;
			if (follow.next != null) {
				follow = follow.next;
			}
		}
		
		follow = pre.next;
		pre.next = null;
		pre = head;
		while (follow != null) {
			stack.push(follow);
			follow = follow.next;
		}
		
		while (!stack.isEmpty()) {
			ListNode temp = stack.pop();
			temp.next = pre.next;
			pre.next = temp;
			
			pre = temp.next;
		}
    }
}