链表反转

leetcode 206 单链表反转

输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL

迭代方法

首先设置pre,cur,lat三个指针

pre   cur  lat
null   1 -> 2 -> 3 -> 4 -> 5 -> null

接着cur.next = pre

pre   cur  lat
null <-1    2 -> 3 -> 4 -> 5 -> null

接着pre = cur，cur = lat，lat = lat.next

      pre  cur  lat
null <-1    2 -> 3 -> 4 -> 5 -> null

重复上述操作直到lat=None。

                     pre  cur  lat
null <-1 <- 2 <- 3 <- 4    5 -> null

代码

def reverseList(self, head: ListNode) -> ListNode:
        pre = None
        cur = head 
        while cur != None:
            cur.next, pre, cur =  pre, cur,cur.next
        return pre

递归

def reverseList(self, head):
    if head == None or head.next == None:
        return head
    node = self.reverseList(head.next)
    head.next.next = head
    head.next = None
    return node

leetcode 25 单链表-k组反转

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