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:
    POJ-2112 Optimal Milking(floyd+最大流+二分)
    网络流之最大流与最小费用流入门&&模板
    0316 校赛训练赛3 解题报告
    string的子串截取
    hash题目大汇总
    Codeforces Round #235 (Div. 2)
    poj2002 -- 点的hash
    hlgHCPC2014校赛训练赛 1 BB.序列问题
    树状数组模板,RMQ模板
    从未放弃--2014.1.21
  • 原文地址:https://www.cnblogs.com/grandyang/p/5657151.html
Copyright © 2011-2022 走看看