leetcode 328. Odd Even Linked List 奇偶序列的调整 + 暴力做法真棒

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:

Given 1->2->3->4->5->NULL,

return 1->3->5->2->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 …

这是一道很简单的链表的操作题，直接遍历一次即可完成。

代码如下：

/* class ListNode
{
int val;
ListNode next;
ListNode(int x) { val = x; }
}*/

/*
* 链表依次遍历即可完成所有的奇数偶数的分离
* */
public class Solution
{
public ListNode oddEvenList(ListNode head)
{
if(head==null || head.next==null)
return head;

ListNode oddHead=new ListNode(-1);
ListNode evenHead=new ListNode(-1);
ListNode iter=head,evenIter=evenHead,oddIter=oddHead;
boolean flag=true;
while(iter!=null)
{
ListNode one=iter;
iter=iter.next;
one.next=null;

if(flag)
{
oddIter.next=one;
oddIter=oddIter.next;
}else
{
evenIter.next=one;
evenIter=evenIter.next;
}
flag = !flag;
}

oddIter.next=evenHead.next;
head=oddHead.next;
return head;
}
}

下面是C++的做法，就这么做吧

代码如下：

#include <iostream>
#include <vector>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <numeric>
#include <cmath>
#include <regex>

using namespace std;

/*
struct ListNode
{
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
*/

class Solution
{
public:
ListNode* oddEvenList(ListNode* head)
{
if (head == NULL || head->next == NULL)
return head;
ListNode* odd = new ListNode(-1);
ListNode* even = new ListNode(-1);
ListNode* i = head, *j = odd, *k = even;
bool flag = true;
while (i != NULL)
{
ListNode* tmp = i;
i = i->next;
tmp->next = NULL;

if (flag == true)
{
j->next = tmp;
j = j->next;
}
else
{
k->next = tmp;
k = k->next;
}
flag = !flag;
}
j->next = even->next;
return odd->next;
}
};