328-m-Odd Even Linked List
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将单链表中所有奇数位和偶数位的节点按顺序平分成两部分,奇数位的在前部分。要求用O(1),否则新建2个表再连起来没意义。
考虑设置一个奇数位指针和一个偶数位指针,分别将链表中奇数节点和偶数节点串联起来,就像串珠子一样,最后将奇数链的末位与偶数链的首位相连即可。
写的时候有个小坑,就是奇数或偶数链走到最后时,要将另一个链表中的末节点的next置为NULL因为此时两个链的末位都是指向该节点,否则连起来后会成为循环链表。。。
AC如下,但是开头还是对特殊情况做了判断,没有做到逻辑泛用。。。
struct ListNode* oddEvenList(struct ListNode* head) { if (head == NULL || head->next == NULL || head->next->next == NULL) return head; struct ListNode *result = head, *odd = head, *even = head->next; struct ListNode *oddhead = head, *evenhead = head->next; while (odd->next != NULL || even->next != NULL) { if (odd->next->next != NULL) { odd->next = odd->next->next; odd = odd->next; } if (odd->next == NULL) { even->next = NULL; break; } if (even->next->next != NULL) { even->next = even->next->next; even = even->next; } if (even->next == NULL) { odd->next = NULL; break; } } odd->next = evenhead; return result;}
后来想了想觉得还可以优化一下,就是每次循环都将奇偶链向后移动,直到其中某一个遇到NULL就退出。这样就将流程统一了不再需要分别两种链是否有NULL。不过AC耗时反而多了2ms。
如下:
struct ListNode* oddEvenList(struct ListNode* head) { if (head == NULL) return head; struct ListNode *result = head, *odd = head, *even = head->next; struct ListNode *oddhead = head, *evenhead = head->next; while (odd->next != NULL && even->next != NULL) { odd->next = odd->next->next; odd = odd->next; even->next = even->next->next; even = even->next; } odd->next = evenhead; return result;}
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