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：

1. 这题不是很难，但是考察的点比较多
2. 将链表一份为二，然后将后半段的链表倒置，然后合并链表，合并的规则是依次从两个链表中取一个
3. 所以这道题考察这些知识点：求链表的中间位置，倒转链表，合并链表

代码如下：

 public void reorderList(ListNode head) {
        if(head == null || head.next == null) return;

        ListNode mid = getMid(head);
        ListNode head2 = mid.next;
        head2 = reverse(head2);
        mid.next = null;
        head = merge(head,head2);
    }


    // 获取链表的中间节点
    private ListNode getMid(ListNode head){
        ListNode pre = head;
        ListNode slow = head;
        ListNode fast = head;

        while(fast != null && fast.next != null){
            pre = slow;
            slow = slow.next;
            fast = fast.next.next;
        }
        return pre;
    }

   // 倒转链表
    private ListNode reverse(ListNode l1){
        ListNode temp = null;
        while(l1 != null){
            ListNode p = l1.next;
            l1.next = temp;
            temp = l1;
            l1 = p;
        }
        return temp;

    }

    // 合并链表
    private ListNode merge(ListNode l1, ListNode l2){
        ListNode dummy = new ListNode(-1);
        ListNode cur = dummy;

        boolean flag = true;
        while(l1 != null && l2 != null){
            if(flag){
                cur.next = l1;
                l1 = l1.next;
                flag = false;
            }else
            {
                cur.next = l2;
                l2 = l2.next;
                flag = true;
            }
            cur = cur.next;
        }

        if(l1 != null) cur.next = l1;
        if(l2 != null) cur.next = l2;

        return dummy.next;
    }

转载于:https://my.oschina.net/happywe/blog/3083411