LeetCode之 Reorder List解决思路

题目如下：

Given a singly linked list L: L₀→L₁→…→L_n-1→L_n,
reorder it to: L₀→L_n→L₁→L_n-1→L₂→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}.

解决思路：

(1)先将原链表分割成两部分。不管原链表长度为偶数或者奇数，第一部分都取长度为 (原链表长度 / 2) +1， 第二部分取剩余的。 分割后的两个链表都要将末尾置为NULL。

这种分割方法，需要先遍历链表长度，再进行分割，需要遍历两次链表。因此可以改进成采用快慢节点取得中间节点的方式，只要一遍遍历即可。

(2)反转链表的第二部分

(3)合并链表的两部分

/** * 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) {       //采用快慢节点找中间节点的方式        if(head == null || head.next == null || head.next.next == null) return;        ListNode Head = split(head);        merge(head, Head);            }        private int length(ListNode head){        int len = 0;        ListNode p = head;        while(p != null){            len++;            p = p.next;        }        return len;    }        private ListNode split(ListNode head, int len){        int splitPos = (len / 2) + 1;        ListNode p = head;        splitPos--;        while(splitPos != 0){            p = p.next;            splitPos--;        }        ListNode Head = p.next;        p.next = null;        return reverse(Head);    } private ListNode split(ListNode head){        ListNode slow = head;        ListNode fast = slow.next;        while(fast.next != null && fast.next.next != null)                    slow = slow.next;            fast = fast.next.next;        }        slow = slow.next;        ListNode Head = slow.next;        slow.next = null;        return reverse(Head);    }        private ListNode reverse(ListNode p){        if(p == null || p.next == null) return p;        ListNode current = p.next;        p.next = null;        while(current != null){            ListNode temp = current;            current = current.next;            temp.next = p;            p = temp;        }        return p;    }        private void merge(ListNode p, ListNode q){        ListNode current = p;        while(q != null){           ListNode temp = q;           q = q.next;           temp.next = current.next;           current.next = temp;           current = current.next.next;        }    }}