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[Leetcode]25. Reverse Nodes in k-Group(C++)

题目描述

题目链接: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.

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.

例子

例子 1

Input: head = [1,2,3,4,5], k = 2
Output: [2,1,4,3,5]

例子 2

Input: head = [1,2,3,4,5], k = 3
Output: [3,2,1,4,5]

例子 3

Input: head = [1,2,3,4,5], k = 1
Output: [1,2,3,4,5]

Follow Up

  • Could you solve the problem in O(1) extra memory space?
  • You may not alter the values in the list’s nodes, only nodes itself may be changed.

Constraints

  • The number of nodes in the list is in the range sz.
  • 1 <= sz <= 5000
  • 0 <= Node.val <= 1000
  • 1 <= k <= sz

解题思路

这道题是 [Leetcode]92. Reverse Linked List II(C++) 的升级版, 92 题要求在一定范围内反转链表,这道题要求每隔 k 个节点翻转链表,思路都是类似的,保存好每 k 个节点的第一个和最后一个节点,做以下操作:

  • 保存前 k 个节点的尾 prev_tail,和当前 k 个节点的头 curr_head
  • 在第 k 个节点 curr_tail 时先保持好下一个节点作为下 k 个节点的头 next_head ,然后断开(curr_tail->next = nullptr
  • 翻转当前 k 个节点 reverse(curr_head)
  • 让前 k 个节点的尾指向保存原先的第 k 个节点(现在是翻转后的 k 个节点的头)prev_tail->next = curr_tail
  • 重置这个节点,对下 k 个节点进行重复操作

代码如下:

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class Solution {
public:
ListNode* reverseKGroup(ListNode* head, int k) {
ListNode* dummy = new ListNode();
ListNode* last_tail = dummy;
ListNode* curr = head;
ListNode* curr_head = curr;

int len = 0;
while (curr) {
len++;
if (len == k) {
// store the node just after next k nodes
ListNode* next_head = curr->next;

// split the linked list
curr->next = nullptr;

// reverse next k nodes
reverse(curr_head);

// the last node of next k nodes should be the head of
// reverse linked list
last_tail->next = curr;

// the head of next k nodes now should be the last of
// next k nodes
last_tail = curr_head;

// reset
curr_head = next_head;
curr = curr_head;
len = 0;
} else {
curr = curr->next;
}
}

last_tail->next = curr_head;

return dummy->next;
}

private:
void reverse(ListNode* head) {
if (!head) return;

ListNode* prev = nullptr;
ListNode* curr = head;

while (curr) {
ListNode* next = curr->next;
curr->next = prev;
prev = curr;
curr = next;
}
}
};
  • 时间复杂度: O(n)
  • 空间复杂度: O(1)

GitHub 代码同步地址: 25.ReverseNodesInKGroup.cpp

其他题目:
GitHub: Leetcode-C++-Solution
博客: Leetcode-Solutions