LeetCode 题解（142）: Reorder List

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题目：

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}.

题解：

做完In-place的sort list，这题就变得非常easy了。先找到中点，如1->2->3->4->5->6中的3，或1->2->3->4->5中的3，将中点之后的链表reverse，可用循环reverse或递归reverse，上一题已给出解法。之后就是sort list中merge的简化版了。

C++版：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    void reorderList(ListNode* head) {        if(head == NULL)            return;                int len = 0;        ListNode* p = head;        while(p != NULL) {            p = p->next;            len++;        }                int i = 0;        p = head;        ListNode* q = p;        while(i < (len - 1) / 2) {            q = q->next;            i++;        }                q->next = reverseTail(q->next);                ListNode* temp = q->next;        q->next = NULL;        q = temp;        while(q != NULL) {            temp = p->next;            p->next = q;            p = temp;            temp = q->next;            q->next = p;            q = temp;        }    }        ListNode* reverseTail(ListNode* head) {        if(head == NULL || head->next == NULL) return head;        ListNode* newHead = reverseTail(head->next);        head->next->next = head;        head->next = NULL;        return newHead;    }};

Java版：

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public void reorderList(ListNode head) {        if(head == null)            return;                    ListNode p = head;        int len = 0;        while(p != null) {            p = p.next;            len++;        }                p = head;        ListNode q = p;        int i = 0;        while(i < (len - 1) / 2) {            q = q.next;            i++;        }                q.next = reverseList(q.next);        ListNode temp = q.next;        q.next = null;        q = temp;        while(q != null) {            temp = p.next;            p.next = q;            p = temp;            temp = q.next;            q.next = p;            q = temp;        }    }        public ListNode reverseList(ListNode head) {        if(head == null)            return null;        ListNode pre = head;        ListNode cur = pre.next;        while(cur != null) {            pre.next = cur.next;            cur.next = head;            head = cur;            cur = pre.next;        }                return head;    }}

Python版：

# Definition for singly-linked list.# class ListNode:#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution:    # @param {ListNode} head    # @return {void} Do not return anything, modify head in-place instead.    def reorderList(self, head):        if head == None:            return                length, p = 0, head        while p != None:            p = p.next            length += 1                    p, q, i = head, head, 0        while i < (length - 1) / 2:            q = q.next            i += 1                    q.next = self.reverseList(q.next)        temp = q.next        q.next = None        q = temp        while q != None:            temp = p.next            p.next = q            p = temp            temp = q.next            q.next = p            q = temp            def reverseList(self, head):        if head == None:            return None                    pre, cur = head, head.next        while cur != None:            pre.next = cur.next            cur.next = head            head = cur            cur = pre.next                    return head

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