zoukankan      html  css  js  c++  java
  • 25. Reverse Nodes in k-Group

    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.

    For 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

    解法:

    采用递归法或者迭代法

    /**
     * Definition for singly-linked list.
     * struct ListNode {
     *     int val;
     *     ListNode *next;
     *     ListNode(int x) : val(x), next(NULL) {}
     * };
     */
    class Solution {
    public:
        ListNode* reverseKGroup(ListNode* head, int k) {
            if(!head || !head->next || k < 2) return head;
            ListNode* next_group = head;
            for(int i =0 ;i < k ; ++i){
                if(next_group) next_group =next_group->next;
                else return head;
            }
            ListNode* next_head = reverseKGroup(next_group,k);
            ListNode *pre = NULL , *cur = head;
            while(cur != next_group) {
                ListNode* next = cur->next;
                cur->next = pre?pre:next_head;
                pre = cur;
                cur = next;
            }
            return pre;
        }
    };
    class Solution {
    public:
        istNode *reverseKGroup(ListNode *head, int k) {
            if (head == nullptr || head->next == nullptr || k < 2) return head;
            ListNode dummy(-1);
            dummy.next = head;
            for(ListNode *prev = &dummy, *end = head; end; end = prev->next) {
                for (int i = 1; i < k && end; i++)
                    end = end->next;
                if (end == nullptr) break; // 不足 k 个
                prev = reverse(prev, prev->next, end);
            }
            return dummy.next;
        }
        // prev 是 first 前一个元素, [begin, end] 闭区间,保证三者都不为 null
        // 返回反转后的倒数第 1 个元素
        ListNode* reverse(ListNode *prev, ListNode *begin, ListNode *end) {
            ListNode *end_next = end->next;
            for (ListNode *p = begin, *cur = p->next, *next = cur->next;
            cur != end_next;p = cur, cur = next, next = next ? next->next : nullptr) {
                cur->next = p;
            }
            begin->next = end_next;
            prev->next = end;
            return begin;
    }
    };
  • 相关阅读:
    uva 1605 building for UN ——yhx
    uva 120 stacks of flapjacks ——yhx
    uva133-S.B.S.
    Uva10082 WERTYU -S.B.S.
    Quicksum-S.B.S.
    NOIP2014提高组 DAY1 -SilverN
    NOIP2013普及组 -SilverN
    uva 1354 Mobile Computing ——yhx
    UVa 11292 Dragon of Loowater
    UVa 839 Not so Mobile
  • 原文地址:https://www.cnblogs.com/CarryPotMan/p/5343684.html
Copyright © 2011-2022 走看看