[LeetCode] 25. k个一组翻转链表

题目链接: https://leetcode-cn.com/problems/reverse-nodes-in-k-group/

题目描述:

给出一个链表，每 k 个节点一组进行翻转，并返回翻转后的链表。

k 是一个正整数，它的值小于或等于链表的长度。如果节点总数不是 k 的整数倍，那么将最后剩余节点保持原有顺序。

示例:

示例 :

给定这个链表：1->2->3->4->5

当 k = 2 时，应当返回: 2->1->4->3->5

当 k = 3 时，应当返回: 3->2->1->4->5

说明 :

• 你的算法只能使用常数的额外空间。
• 你不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。

思路:

思路一:

用栈,我们把k个数压入栈中,然后弹出来的顺序就是翻转的!

这里要注意几个问题

第一,剩下的链表个数够不够k个(因为不够k个不用翻转);

第二,已经翻转的部分要与剩下链表连接起来

思路二:

尾插法

直接举个例子: k = 3

pre
tail    head
dummy    1     2     3     4     5
# 我们用tail 移到要翻转的部分最后一个元素
pre     head       tail
dummy    1     2     3     4     5
		cur
# 我们尾插法的意思就是,依次把cur移到tail后面
pre          tail  head
dummy    2     3    1     4     5
		cur
# 依次类推
pre     tail      head
dummy    3     2    1     4     5
		cur
....

思路3:

递归

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代码:

思路1:

栈

python

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

class Solution:
    def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
        dummy = ListNode(0)
        p = dummy
        while True:
            count = k
            stack = []
            tmp = head
            while count and tmp:
                stack.append(tmp)
                tmp = tmp.next
                count -= 1
            # 注意,目前tmp所在k+1位置
            # 说明剩下的链表不够k个,跳出循环
            if count :
                p.next = head
                break
            # 翻转操作
            while stack:
                p.next = stack.pop()
                p = p.next
            #与剩下链表连接起来
            p.next = tmp
            head = tmp

        return dummy.next

java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        Deque<ListNode> stack = new ArrayDeque<ListNode>();
        ListNode dummy = new ListNode(0);
        ListNode p = dummy;
        while (true) {
            int count = 0;
            ListNode tmp = head;
            while (tmp != null && count < k) {
                stack.add(tmp);
                tmp = tmp.next;
                count++;
            }
            if (count != k) {
                p.next = head;
                break;
            }
            while (!stack.isEmpty()){
                p.next = stack.pollLast();
                p = p.next;
            }
            p.next = tmp;
            head = tmp;
        }
        return dummy.next;
    }
}

思路2:

尾插法

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

class Solution:
    def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
        dummy = ListNode(0)
        dummy.next = head
        pre = dummy
        tail = dummy
        while True:
            count = k
            while count and tail:
                count -= 1
                tail = tail.next
            if not tail: break
            head = pre.next
            while pre.next != tail:
                cur = pre.next # 获取下一个元素
                # pre与cur.next连接起来,此时cur(孤单)掉了出来
                pre.next = cur.next
                cur.next = tail.next # 和剩余的链表连接起来
                tail.next = cur #插在tail后面
            # 改变 pre tail 的值
            pre = head
            tail = head
        return dummy.next

java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode pre = dummy;
        ListNode tail = dummy;
        while (true) {
            int count = 0;
            while (tail != null && count != k) {
                count++;
                tail = tail.next;
            }
            if (tail == null) break;
            ListNode head1 = pre.next;
            while (pre.next != tail) {
                ListNode cur = pre.next;
                pre.next = cur.next;
                cur.next = tail.next;
                tail.next = cur;
            }
            pre = head1;
            tail = head1;
        }
        return dummy.next;
    }
}

思路3:

递归

python

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

class Solution:
    def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
        cur = head
        count = 0
        while cur and count!= k:
            cur = cur.next
            count += 1
        if count == k:
            cur = self.reverseKGroup(cur, k)
            while count:
                tmp = head.next
                head.next = cur
                cur = head
                head = tmp
                count -= 1
            head = cur
        return head

java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode cur = head;
        int count = 0;
        while (cur != null && count != k) {
            cur = cur.next;
            count++;
        }
        if (count == k) {
            cur = reverseKGroup(cur, k);
            while (count != 0) {
                count--;
                ListNode tmp = head.next;
                head.next = cur;
                cur = head;
                head = tmp;
            }
            head = cur;
        }
        return head;
    }