leetcode——25. K 个一组翻转链表

用栈做：

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

class Solution(object):
    def reverseKGroup(self, head, k):
        """
        :type head: ListNode
        :type k: int
        :rtype: 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
            if count:
                p.next=head
                break
            while stack:
                p.next=stack.pop()
                p=p.next
            p.next=tmp
            head=tmp
        return dummy.next

执行用时 :32 ms, 在所有 python 提交中击败了95.97%的用户

内存消耗 :13.2 MB, 在所有 python 提交中击败了44.26%的用户

 

 

——2019.10.31

 

---

public ListNode reverseKGroup(ListNode head, int k) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;

        ListNode pre = dummy;
        ListNode end = dummy;

        while (end.next != null) {
            for (int i = 0; i < k && end != null; i++) end = end.next;
            if (end == null) break;
            ListNode start = pre.next;
            ListNode next = end.next;
            end.next = null;
            pre.next = reverse(start);
            start.next = next;
            pre = start;

            end = pre;
        }
        return dummy.next;
    }

    private ListNode reverse(ListNode head) {
        ListNode pre = null;
        ListNode curr = head;
        while (curr != null) {
            ListNode next = curr.next;
            curr.next = pre;
            pre = curr;
            curr = next;
        }
        return pre;
    }

 自己还是没能做出来。

——2020.7.14