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LeetCode 算法题 25. Reverse Nodes in k-Group

分类:algorithm| 发布时间:2016-07-17 15:36:00


题目

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

题意

给出一个链表,每次以 k 为一组反转节点。只能修改节点本身,不能修改节点中的值。

解法

/**
 * 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) {
        ListNode dummy(0);
        dummy.next = head;
        ListNode *tail = &dummy;
        stack<ListNode*> s;
        while (head) {
            int i = 0;
            for (; head && i < k; ++i) {
                s.push(head);
                head = head->next;
            }

            if (i == k) {
                for (; i > 0; --i) {
                    tail->next = s.top();
                    tail = tail->next;
                    s.pop();
                }

                tail->next = head;
            }
        }

        return dummy.next;
    }
};