LeetCode刷题笔记（Odd Even Linked List）

趁热打铁，又刷了一道中档题，感觉难度还行，下面就和大家来分享一下经验吧！

题目如下：

Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.

You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.

Example 1:

Input: 1->2->3->4->5->NULL
Output: 1->3->5->2->4->NULL
Example 2:

Input: 2->1->3->5->6->4->7->NULL
Output: 2->3->6->7->1->5->4->NULL
Note:

The relative order inside both the even and odd groups should remain as it was in the input.
The first node is considered odd, the second node even and so on ...

题意分析：

给定一个单链表，将链表中所有的奇数节点放在偶数节点之前，且该程序应该是O（1）的空间复杂度和O（节点个数）的时间复杂度。

注：① 应该保证原始节点的相对顺序不变。

② 第一个节点被看做是奇数节点，第二个节点被看做是偶数节点，以此类推。

解答如下：

方法一（新建链表法）

与（https://mp.csdn.net/postedit/88654310）中方法二的思想一样，所以只需在其基础上进行稍加修改便可求解。

class Solution{
public:
    ListNode* oddEvenList(ListNode* head){
        ListNode* dummy = new ListNode( -1 );
        dummy -> next = head;
        ListNode* cur = dummy;
        ListNode* newdummy = new ListNode( -1 );
        ListNode* p = newdummy;
        int count = 1;                                //记录奇偶节点
        while ( cur -> next ){
            if ( count % 2 != 0)                      //判断奇偶节点，并将奇节点放入新链表中
            {
                p -> next = cur -> next;
                p = p -> next;
                cur -> next = cur -> next -> next;
                p -> next = NULL;
                count ++;
            }
            else
            {cur = cur -> next;
                count ++;}
        }
        p -> next = dummy -> next;                    //将偶节点链表接在奇节点链表后
        return newdummy -> next;
    }

};

提交后的结果如下：

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方法二（节点插入法）

新建两个指针，让pre指向奇节点，cur指向偶节点，然后将cur后面的那个奇节点插入到pre的后面，然后pre和cur各自向前移动一位，此时cur又会指向偶节点，pre指向当前奇节点的末尾。直至循环结束，可以得到一个奇节点在前偶节点在后的链表。

class Solution{
public:
    ListNode* oddEvenList( ListNode* head){
        if ( !head || !head -> next) return head;
        ListNode* pre = head;
        ListNode* cur = head -> next;

        while( cur && cur -> next ){
            ListNode * tmp = pre -> next;
            pre -> next = cur -> next;
            cur -> next = cur -> next -> next;
            pre -> next -> next = tmp;
            cur = cur -> next;
            pre = pre -> next;
        }
        return head;
    }

};

提交后的结果如下：

日积月累，与君共进，增增小结，未完待续。

LeetCode刷题笔记（Odd Even Linked List）

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