zoukankan      html  css  js  c++  java
  • [LintCode] Reverse Nodes in k-Group 每k个一组翻转链表

    Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

    If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

    You may not alter the values in the nodes, only nodes itself may be changed.
    Only constant memory is allowed.

    Example

    Given this linked list: 1->2->3->4->5

    For k = 2, you should return: 2->1->4->3->5

    For k = 3, you should return: 3->2->1->4->5

    LeetCode上的原题,请参见我之前的博客Reverse Nodes in k-Group 每k个一组翻转链表

    解法一:

    class Solution {
    public:
        /**
         * @param head a ListNode
         * @param k an integer
         * @return a ListNode
         */
        ListNode *reverseKGroup(ListNode *head, int k) {
            ListNode *dummy = new ListNode(-1), *pre = dummy, *cur = pre;
            dummy->next = head;
            while (cur->next) {
                int d = k;
                while (cur->next && d-- > 0) {
                    cur = cur->next;
                }
                if (d > 0) return dummy->next;
                ListNode *t = cur->next;
                cur->next = NULL;
                pre->next = reverse(pre->next);
                for (int i = 0; i < k; ++i) pre = pre->next;
                pre->next = t;
                cur = pre;
            }
            return dummy->next;
        }
        ListNode* reverse(ListNode *head) {
            ListNode *dummy = new ListNode(-1), *cur = head;
            dummy->next = head;
            while (cur && cur->next) {
                ListNode *t = cur->next;
                cur->next = t->next;
                t->next = dummy->next;
                dummy->next = t;
            }
            return dummy->next;
        }
    };

    解法二:

    class Solution {
    public:
        /**
         * @param head a ListNode
         * @param k an integer
         * @return a ListNode
         */
        ListNode *reverseKGroup(ListNode *head, int k) {
            ListNode *dummy = new ListNode(-1), *pre = dummy, *cur = pre;
            dummy->next = head;
            int num = 0;
            while (cur = cur->next) ++num;
            while (num >= k) {
                cur = pre->next;
                for (int i = 1; i < k; ++i) {
                    ListNode *t = cur->next;
                    cur->next = t->next;
                    t->next = pre->next;
                    pre->next = t;
                }
                pre = cur;
                num -= k;
            }
            return dummy->next;
        }
    };

    解法三:

    class Solution {
    public:
        /**
         * @param head a ListNode
         * @param k an integer
         * @return a ListNode
         */
        ListNode *reverseKGroup(ListNode *head, int k) {
            ListNode *cur = head;
            for (int i = 0; i < k; ++i) {
                if (!cur) return head;
                cur = cur->next;
            }
            ListNode *new_head = reverse(head, cur);
            head->next = reverseKGroup(cur, k);
            return new_head;
        }
        ListNode* reverse(ListNode* head, ListNode* tail) {
            ListNode *pre = tail;
            while (head != tail) {
                ListNode *t = head->next;
                head->next = pre;
                pre = head;
                head = t;
            }
            return pre;
        }
    };
  • 相关阅读:
    LocalSessionFactoryBean有几个属性查找hibernate映射文件
    关于Spring中配置LocalSessionFactoryBean来生成SessionFactory
    【Spring源码分析】配置文件读取流程
    Java序列化接口的作用总结1
    Java序列化接口的作用总结
    hibernate抓取策略
    170531、FormData 对象的使用
    170530、java 迭代hashmap常用的三种方法
    170529、springMVC 的工作原理和机制
    170526、spring 执行定时任务
  • 原文地址:https://www.cnblogs.com/grandyang/p/5657151.html
Copyright © 2011-2022 走看看