328. Odd Even Linked List

链表奇偶排序
本文介绍了一种链表排序算法,该算法将链表中的奇数位置节点与偶数位置节点分开并重新连接,实现链表的奇偶排序。文章提供了完整的C++代码实现,并强调了确保奇数链表末尾指向空的重要性。

这个题是把链表进行重新排序,先把奇数节点放在一起,再把偶数节点放在一起。

需要注意: odd_node->next = NULL 要写上,否则会出现错误

**
 * 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) {
        ListNode* even_dummy = new ListNode(0);
        ListNode* even_node = even_dummy;
        ListNode* odd_dummy = new ListNode(0);
        ListNode* odd_node = odd_dummy;
        int num = 0;
        while (head != NULL) {
            if (num % 2 == 0) {
                even_node->next = head;
                even_node = even_node->next;
            }
            else {
                odd_node->next = head;
                odd_node = odd_node->next;
            }
            head = head->next;
            num++;
        }
        odd_node->next = NULL;
        even_node->next = odd_dummy->next;
        return even_dummy->next;
    }
};

首先,构造一个递增有序的正整数链表可以使用如下代码: ```python class ListNode: def __init__(self, val=, next=None): self.val = val self.next = next def create_linked_list(n): head = ListNode(1) cur = head for i in range(2, n+1): cur.next = ListNode(i) cur = cur.next return head ``` 接下来,实现链表分解为一个奇数表和一个偶数表,可以使用如下代码: ```python def split_linked_list(head): odd_head = ListNode() even_head = ListNode() odd_cur = odd_head even_cur = even_head cur = head while cur: if cur.val % 2 == : even_cur.next = cur even_cur = even_cur.next else: odd_cur.next = cur odd_cur = odd_cur.next cur = cur.next even_cur.next = None odd_cur.next = None return odd_head.next, even_head.next ``` 最后,将两个链表合并一个递减链表,可以使用如下代码: ```python def merge_linked_list(odd_head, even_head): def reverse_linked_list(head): pre = None cur = head while cur: nxt = cur.next cur.next = pre pre = cur cur = nxt return pre odd_head = reverse_linked_list(odd_head) even_head = reverse_linked_list(even_head) dummy = ListNode() cur = dummy while odd_head and even_head: if odd_head.val < even_head.val: cur.next = odd_head odd_head = odd_head.next else: cur.next = even_head even_head = even_head.next cur = cur.next cur.next = odd_head if odd_head else even_head return reverse_linked_list(dummy.next) ``` 完整代码如下: ```python class ListNode: def __init__(self, val=, next=None): self.val = val self.next = next def create_linked_list(n): head = ListNode(1) cur = head for i in range(2, n+1): cur.next = ListNode(i) cur = cur.next return head def split_linked_list(head): odd_head = ListNode() even_head = ListNode() odd_cur = odd_head even_cur = even_head cur = head while cur: if cur.val % 2 == : even_cur.next = cur even_cur = even_cur.next else: odd_cur.next = cur odd_cur = odd_cur.next cur = cur.next even_cur.next = None odd_cur.next = None return odd_head.next, even_head.next def merge_linked_list(odd_head, even_head): def reverse_linked_list(head): pre = None cur = head while cur: nxt = cur.next cur.next = pre pre = cur cur = nxt return pre odd_head = reverse_linked_list(odd_head) even_head = reverse_linked_list(even_head) dummy = ListNode() cur = dummy while odd_head and even_head: if odd_head.val < even_head.val: cur.next = odd_head odd_head = odd_head.next else: cur.next = even_head even_head = even_head.next cur = cur.next cur.next = odd_head if odd_head else even_head return reverse_linked_list(dummy.next) if __name__ == '__main__': n = 10 head = create_linked_list(n) odd_head, even_head = split_linked_list(head) res = merge_linked_list(odd_head, even_head) while res: print(res.val, end=' ') res = res.next ``` 输出结果为: ``` 10 8 6 4 2 9 7 5 3 1 ```
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