Reorder List 微软笔试题

最新推荐文章于 2014-08-31 21:32:35 发布

转载最新推荐文章于 2014-08-31 21:32:35 发布 · 51 阅读

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原文链接：http://www.cnblogs.com/727713-chuan/p/3409042.html

本文介绍如何使用一种方法在不改变节点值的情况下，重新排列单链表的顺序，具体步骤包括反转链表后半部分并重新插入。

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}

现在在Leetcode上也有了，这题的做法有很多，我是用的是一个舍友给我提供的一种方法，先把后面部分反转在重新插入即可。

/**
 * 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) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int length;
        length = 0;
        ListNode *p,*q,*t;
        p = head;
        while(p)
        {
            p = p->next;
            length++;
        }
        if(length<=2)
        return ;
        p = head;
        length = length/2;
        for(int i = 0;i<length;i++)
        {
            p = p->next;
        }
        t = p->next;
        p->next = NULL;
        p = t;
        t = t->next;
        p->next = NULL;
        while(t)
        {
            q = t->next;
            t->next = p;
            p = t;
            t = q;
        }
        t = head;
        while(p&&t)
        {
            q = p->next;
            p->next = t->next;
            t->next = p;
            t = t->next->next;
            p = q;
        }
    }
};

转载于:https://www.cnblogs.com/727713-chuan/p/3409042.html