[LeetCode] 25. 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.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

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

Note:

• Only constant extra memory is allowed.
• may not alter the values in the list‘s nodes, only nodes itself may be changed.

给一个链表和正整数k，以每k个为一组来翻转链表。把原链表分成若干小段，然后分别对其进行翻转。

解法1：迭代

解法2: 递归

Java:

public ListNode reverseKGroup(ListNode head, int k) {
    ListNode curr = head;
    int count = 0;
    while (curr != null && count != k) { // find the k+1 node
        curr = curr.next;
        count++;
    }
    if (count == k) { // if k+1 node is found
        curr = reverseKGroup(curr, k); // reverse list with k+1 node as head
        // head - head-pointer to direct part,
        // curr - head-pointer to reversed part;
        while (count-- > 0) { // reverse current k-group:
            ListNode tmp = head.next; // tmp - next head in direct part
            head.next = curr; // preappending "direct" head to the reversed list
            curr = head; // move head of reversed part to a new node
            head = tmp; // move "direct" head to the next node in direct part
        }
        head = curr;
    }
    return head;
}

Python:

# Definition for singly-linked list.
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

    def __repr__(self):
        if self:
            return "{} -> {}".format(self.val, repr(self.next))

class Solution:
    # @param head, a ListNode
    # @param k, an integer
    # @return a ListNode
    def reverseKGroup(self, head, k):
        dummy = ListNode(-1)
        dummy.next = head

        cur, cur_dummy = head, dummy
        length = 0

        while cur:
            next_cur = cur.next
            length = (length + 1) % k

            if length == 0:
                next_dummy = cur_dummy.next
                self.reverse(cur_dummy, cur.next)
                cur_dummy = next_dummy

            cur = next_cur

        return dummy.next

    def reverse(self, begin, end):
            first = begin.next
            cur = first.next

            while cur != end:
                first.next = cur.next
                cur.next = begin.next
                begin.next = cur
                cur = first.next

C++:

// Time:  O(n)
// Space: O(1)

/**
 * 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) {
        ListNode dummy{0};
        dummy.next = head;
        auto curr = head, curr_dummy = &dummy;
        int len = 0;

        while (curr) {
            auto next_curr = curr->next;
            len = (len + 1) % k;

            if (len == 0) {
                auto next_dummy = curr_dummy->next;
                reverse(&curr_dummy, curr->next);
                curr_dummy = next_dummy;
            }
            curr = next_curr;
        }
        return dummy.next;
    }

    void reverse(ListNode **begin, const ListNode *end) {
        ListNode *first = (*begin)->next;
        ListNode *curr = first->next;

        while (curr != end) {
            first->next = curr->next;
            curr->next = (*begin)->next;
            (*begin)->next = curr;
            curr = first->next;
        }
    }
};　　

C++:

class Solution {
public:
    ListNode *reverseKGroup(ListNode *head, int k) {
        if (!head || k == 1) return head;
        ListNode *dummy = new ListNode(-1);
        ListNode *pre = dummy, *cur = head;
        dummy->next = head;
        int i = 0;
        while (cur) {
            ++i;
            if (i % k == 0) {
                pre = reverseOneGroup(pre, cur->next);
                cur = pre->next;
            } else {
                cur = cur->next;
            }
        }
        return dummy->next;
    }
    ListNode *reverseOneGroup(ListNode *pre, ListNode *next) {
        ListNode *last = pre->next;
        ListNode *cur = last->next;
        while(cur != next) {
            last->next = cur->next;
            cur->next = pre->next;
            pre->next = cur;
            cur = last->next;
        }
        return last;
    }
};　　

C++:

class Solution {
public:
    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;
    }
};

C++: 递归

class Solution {
public:
    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;
    }
};

类似题目：

[LeetCode] 24. Swap Nodes in Pairs 成对交换节点

原文地址：https://www.cnblogs.com/lightwindy/p/9684918.html