Leetcode--Reorder List

Leetcode--Reorder List

Problem Description:

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：按照要求交換鏈表的元素，第一種方法，直接遍歷一遍鏈表將所有節點指針保存在vector中，然後一頭一尾不斷重建鏈表即可。具體代碼如下：

/**
 * 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) {
        ListNode *p=head;
        int length,i=1,tag=0;
        vector array;
        while(p)
        {
            array.push_back(p);
            p=p->next;
        }
        p=head;
        if(array.size()<3) return;
        length=array.size()-1;
        while(tagnext=array[length--];
            p=p->next;
            p->next=array[i++];
            p=p->next;
            tag++;
        }
        p->next=NULL;
    }
};

第二種方法就是找到鏈表的中間節點，將鏈表分為兩半，然後將後半部分轉置，最後將前後兩部分進行合並即可，具體代碼如下：

/**
 * 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){return;}
        if (head->next==NULL){return;}
        ListNode *p=head; 
        ListNode *q=head; 
         
        //find the midddle pointer
        while (q->next && q->next->next){
            p=p->next;
            q=q->next->next;
        }
         
        //now p is middle pointer
        //reverse p->next to end
        q = p->next;
        while (q->next){
            ListNode* tmp = p->next;
            p->next = q->next;
            q->next = q->next->next;
            p->next->next = tmp;
        }
         
        //reorder
        q = head;
        while (p!=q && p->next){
            ListNode* tmp = q->next;
            q->next = p->next;
            p->next = p->next->next;
            q->next->next = tmp;
            q=q->next->next;
        }
        return;
    }
};