剑指Offer - Python

014-链表中倒数第k个结点

用快慢指针：p2比p1先走k-1（1到k：间隔了k-1）步，然后再一起走，当p2为最后一个时，p1就为倒数第k个数

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def FindKthToTail(self, head, k):
        if not head or k<=0: return None
        p1 = head
        p2 = head
         #设置两个指针，p2指针先走（k-1）步，然后再一起走
        while (k-1)>0:
            if (p2.next != None):
                p2 = p2.next
                k -= 1
            else:
                return None
        while (p2.next != None): # 等同while p2.next:
            p1 = p1.next
            p2 = p2.next
        return p1

015-反转链表（链表转化赋值）

思路：变化node.next

假设翻转1->2->3->4->5，（54321）

重要：

• 1.构建辅助节点head
• 2.我们将p的next指向tp的next
• 3.将tp的next指向head的next
• 4.将head的next指向tp

class ListNode:
     def __init__(self, x):
            self.val = x
            self.next = None

class Solution:
    # 新链表的表头,不是head，是head.next
    def ReverseList(self, pHead):
        if not pHead:return None
        #构建辅助节点head
        head = ListNode(0)
        head.next = pHead
        p = pHead
        while p.next != None:
            #链表节点转化操作（重要）
            tp = p.next
            p.next = tp.next
            tp.next = head.next
            head.next = tp
        return head.next