标签：16px ref ide 链表 list tco 思路 color str
题目链接：https://leetcode.com/problems/odd-even-linked-list/description/
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->NULLOutput:1->3->5->2->4->NULL
Example 2:
Input: 2->1->3->5->6->4->7->NULLOutput:2->3->6->7->1->5->4->NULL
Note:
思路：
编码如下：
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode* oddEvenList(ListNode* head) { 12 if (head == nullptr) return nullptr; 13 14 ListNode *oddHead = nullptr; 15 ListNode *pOdd = nullptr; 16 ListNode *evenHead = nullptr; 17 ListNode *pEven = nullptr; 18 19 int i = 0; 20 ListNode *cur = head; 21 while (nullptr != cur) 22 { 23 i++; 24 ListNode *p = cur->next; 25 if ( i & 1 == 1) // i为奇数时 26 { 27 if (nullptr == oddHead) 28 { 29 oddHead = cur; 30 pOdd = oddHead; 31 pOdd->next = nullptr; 32 } 33 else 34 { 35 pOdd->next = cur; 36 pOdd = pOdd->next; 37 pOdd->next = nullptr; 38 } 39 } 40 else 41 { 42 if (nullptr == evenHead) 43 { 44 evenHead = cur; 45 pEven = evenHead; 46 pEven->next = nullptr; 47 } 48 else 49 { 50 pEven->next = cur; 51 pEven = pEven->next; 52 pEven->next = nullptr; 53 } 54 } 55 56 cur = p; 57 } 58 59 pOdd->next = evenHead; 60 61 return head; 62 } 63 };
标签：16px ref ide 链表 list tco 思路 color str
原文地址：https://www.cnblogs.com/ming-1012/p/9949644.html
