链表_leetcode92

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

class Solution(object):
    def reverseBetween(self, head, m, n):
        """
        :type head: ListNode
        :type m: int
        :type n: int
        :rtype: ListNode
        """

        if not head or not head.next:
            return head
        if n < 1 or m >= n :
            return head

        cout = 1
        cur = head



        # 还是边界条件没考虑完全呀，比如翻转整个链表

        preStart = None
        nextEnd = None

        while cur:

            cur = cur.next

            cout += 1

            if cout == m-1:
                preStart = cur
            if cout == n:
                nextEnd = cur.next

                break


        start = preStart.next

        begin = start.next

        for i in range(m-n):

            nextBegin = begin.next
            begin.next = preStart.next
            preStart.next = begin
            begin = nextBegin

        start.next = nextEnd

        return head



class Solution2(object):
    def reverseBetween(self, head, m, n):
        """
        :type head: ListNode
        :type m: int
        :type n: int
        :rtype: ListNode
        """

        count = 1
        root = ListNode(0)
        root.next = head

        pre = root

        # while pre.next:
        #
        #     pre = pre.next
        #     count += 1
        #
        #     if count == m :
        #         break

        while pre.next and count < m :
            pre = pre.next
            count += 1

        if not pre:
            return head


        mNode = pre.next
        cur = mNode.next

        while cur and count < n:

            next = cur.next
            cur.next = pre.next
            pre.next = cur
            mNode.next = next

            cur = next
            count += 1



        return root.next