leetCodeReorderList链表合并

原题

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}.

分析

将链表结点按照第一个 倒数第一个 第二个 倒数第二个 的顺序重新排序

1 2 3 4重新排序后是 1 4 2 3

AC代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    //翻转链表
    ListNode *reverse(ListNode *head){
        if(!head) return;
        ListNode * pre = NULL;
        while(head){
            ListNode * nex = head->next;
            head->next = pre;
            pre = head;
            head = nex;
        }
        return pre;
    }
    void reorderList(ListNode* head) {
        if(!head || !head->next) return head;
        ListNode * fast = head;
        ListNode * slow = head;
        while(fast->next && fast->next->next){
            fast = fast->next->next;
            slow = slow->next;
        }
        fast = slow->next;
        //截断
        slow->next = NULL;
        //翻转后半段
        ListNode * p = reverse(fast);

        //合并
        ListNode * res = head;
        ListNode * q = head->next;
        while(p && q){
            res->next = p;
            p = p->next;
            res = res->next;

            res->next = q;
            q = q->next;
            res = res->next;
        }
        if(p) res->next = p;
        if(q) res->next = q;
    }
};
posted @ 2017-07-31 18:04  lepeCoder  阅读(204)  评论(0编辑  收藏  举报