题解 | #链表内指定区间反转#

• 没注解

class Solution:
    def reverseBetween(self , head: ListNode, m: int, n: int) -> ListNode:
        # write code here
        pre=None
        cur=head
        
        if m==n:
            return head
          
        for i in range(m-1):
            pre = cur
            cur = cur.next
        m_pre = pre
        m_cur = cur
        
        for i in range(n-m+1):
            tnext = cur.next
            cur.next= pre
            pre=cur
            cur=tnext
            
        if m_pre:
            m_pre.next = pre
        else:
            head = pre
        m_cur.next = cur
        
        return head

• 有注解

class Solution:
    def reverseBetween(self , head: ListNode, m: int, n: int) -> ListNode:
        # write code here
        pre=None
        cur=head
        #   1->2->3->4->5
        # p c
        
        # 当m==n时,不需要反转
        if m==n:
            return head
          
        # 让cur移到m位置下
        for i in range(m-1):
            pre = cur
            cur = cur.next
        m_pre = pre
        m_cur = cur
        # 此时情况
        # 1->2->3->4->5
        # p  c
        # mp mc (为了表示方便,这里的mp时变量m_pre, mc是变量m_cur)
        
        # 反转开始
        for i in range(n-m+1):
            tnext = cur.next
            cur.next= pre
            pre=cur
            cur=tnext
        # 反转结束, 此时情况
        # 1<-2<-3<-4  5
        # mp mc    p  c
        
        # 微调
        if m_pre:
            m_pre.next = pre
        else:
            #当m=1时, m_pre是None, 没有.next属性, 
            head = pre
        m_cur.next = cur
        # 调整结束此时情况
        # 1->4  5<-2<-3<-4 即1 4 3 2 5
        
        return head