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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.
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重排链表,链表的奇数元素都排在偶数元素前面,
而且奇数间的相对位置不变,偶数间的相对位置不变.
思路:
两个假的dummy节点,分别按照 尾插法链接原始,
最后将偶数链表接在奇数链表的后边即可.
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: /// ListNode* oddEvenList(ListNode* head) { if(head==nullptr || head->next==nullptr) return head; ListNode dummy1(-1); ListNode dummy2(-2); ListNode *h1 = &dummy1; ListNode *h2 = &dummy2; int k = 1; while(head){ if(k%2==0){ h1->next = head; h1 = h1->next; }else{ h2->next = head; h2 = h2->next; } k++; head = head->next; } h2->next = dummy1.next; h1->next = nullptr; return dummy2.next; } };
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原文地址：http://www.cnblogs.com/li-daphne/p/5606883.html
