Posts Tagged 【List 】Reorder List

Reorder List

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Question Solution 

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}.

/** * Definition for singly-linked list. * class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */  /* 我的思路比较简单，牺牲内存来处理，明天想想直接用指针来 建立一个headnew，方便处理 最后的next为null */public class Solution {    public void reorderList(ListNode head) {        if(head == null) return;        List<ListNode> list = new ArrayList<ListNode>();        ListNode headNew = new ListNode(0);        headNew.next = head;        ListNode p = head;        while(p!=null) {            list.add(p);            p = p.next;        }        int len = list.size();        p = headNew;        for(int i = 0;i<len/2;i++) {            p.next = list.get(i);            p.next.next = list.get(len-1-i);            p = p.next.next;        }        p.next = list.get(len/2);        p.next.next = null;                }}

</pre><pre code_snippet_id="651774" snippet_file_name="blog_20150423_1_2185049" name="code" class="java">/** * Definition for singly-linked list. * class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */  /* 接着来 将list分成前后两个部分 后一个部分反转 两个部分插入 注意将最后的node的next赋值为null */public class Solution {    public void reorderList(ListNode head) {        if(head == null || head.next == null) return;        twistList(head,reverseList(midList(head)));    }        public ListNode midList(ListNode head) {        if(head == null || head.next == null) return null;        ListNode p = head,q = head.next;        while(q != null && q.next != null) {            p = p.next;            q = q.next.next;        }        return p.next;    }        public ListNode reverseList(ListNode head) {        ListNode p=head,q=head.next;        head.next = null;        while(q != null) {            ListNode temp = q.next;            q.next = p;            p = q;            q = temp;        }        return p;    }        public ListNode twistList(ListNode head1,ListNode head2) {        ListNode q = head1,p = head2;        ListNode newHead = new ListNode(0);        ListNode qp = newHead;        while(q != null) {            if(p != null) {                ListNode qTemp = q.next;                ListNode pTemp = p.next;                qp.next = q;                qp.next.next = p;                qp = p;                q = qTemp;                p = pTemp;               } else {                qp.next = q;                qp.next.next = null;                break;            }                    }        return newHead.next;    }}

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