LeetCode Odd Even Linked List
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 …
链表题，要求把链表中的奇数位节点连接起来，偶数位节点也连接起来，最后这两个链表连接起来。要求O(1)的空间复杂度和O(n)的时间复杂度。 因为奇数链表和偶数链表都是从原链表中隔一个取一个，立马联想到快慢指针，在这个题里，我们设置两个快指针fast1和fast2，分别指向奇数节点和偶数节点，然后分别取下来，接到odd和even链表中。最后把odd的tail指向even的head就搞定了。符合O(1)空间复杂度和O(n)时间复杂度。 完整代码如下： [cpp] class Solution { public: ListNode* oddEvenList(ListNode* head) { if (head == NULL)return head; ListNode *odd = new ListNode(0), *even = new ListNode(0); ListNode *otail = odd, *etail = even; ListNode *fast1 = head, *fast2 = head->next; while (fast1 || fast2) { otail->next = fast1; etail->next = fast2; if (otail)otail = otail->next; if (etail)etail = etail->next; if (fast2)fast1 = fast2->next; else fast1 = NULL; if (fast1)fast2 = fast1->next; else fast2 = NULL; } otail->next = even->next; return odd->next; } }; [/cpp] 本代码提交AC，用时16MS。 二刷。其实不用快慢指针，直接根据奇偶性就可以了，更简洁的代码如下： [cpp] class Solution { public: ListNode* oddEvenList(ListNode* head) { ListNode *odd = new ListNode(0), *even = new ListNode(0); int cnt = 1; ListNode *cur = head, *otail = odd, *etail = even; while(cur) { if(cnt & 1) { otail->next = cur; otail = otail->next; } else { etail->next =cur; etail = etail->next; } cur = cur->next; ++cnt; } otail->next = even->next; etail->next = NULL; // caution!!! return odd->next; } }; [/cpp] 本代码提交AC，用时19MS。]]>