链表重组算法-优快云博客

本文链接：https://blog.youkuaiyun.com/MacanS/article/details/52661780

题目描述

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}.

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代码

语言：C++ 运行时间： <1 ms 占用内存：9740K 状态：答案正确

/**

 *
 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 || !head->next) return

;

        vector<int>
 v;

        ListNode*
 p = head;

        while(p){

            v.push_back(p->val);

            p
 = p->next;

        }

        p
 = head;

        int

i = 0,j
 = v.size()-1,index
 = 0;

        while

(i<=j){

            if(index%2==0){

                p->val
 = v[i++];

            }else{

                p->val
 = v[j--];

            }

            p
 = p->next;

            ++index;

        }

    }

};