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  • LeetCode Solutions : Reorder List

    Given a singly linked list L: L0→L1→…→Ln-1→Ln,
    reorder it to: L
    0→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 itto {1,4,2,3}.

    Considering the following steps:

     * 1. split such list into two list, first and second, according to slow and fast point
     * 2. reverse the second list
     * 3. insert the second list into the first list

    coding solution:


    /**
     * 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){
    			ListNode low=head;//1. split such list into two list, first and second, according to slow and fast point
    			ListNode fast=head;
    			while(fast.next!=null&&fast.next.next!=null){
    				low=low.next;
    				fast=fast.next.next;
    			}
    			ListNode first=head;
    			ListNode second=low.next;
    			low.next=null;
    			
    		    second=reverse(second);//2. reverse the second list
    			while(second!=null){//3. insert the second list into the first list
    				ListNode p1=first.next;
    				ListNode p2=second.next;
    				first.next=second;
    				second.next=p1;
    				first=p1;
    				second=p2;
    			}
    		
    		}
        }
    	private ListNode reverse(ListNode head){
    		if(head==null||head.next==null)
    			return head;
    		ListNode pre=head;
    		ListNode cur=head.next;
    		while(cur!=null){
    			ListNode nextNode=cur.next;
    			cur.next=pre;
    			pre=cur;
    			cur=nextNode;			
    		}
    		head.next=null;
    		return pre;
    	}
    }


    版权声明:本文博主原创文章,博客,未经同意不得转载。

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  • 原文地址:https://www.cnblogs.com/zfyouxi/p/4850687.html
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