Python3-[42]面试题52.两个链表的第一个公共节点

题目

编写一个程序，找到两个单链表相交的起始节点。
如下面的两个链表：

在节点 c1 开始相交。
P.S.题目来源于leetcode

解题思路

解法1:hash map

使用一个集合存储A的所有节点，再判断B的节点是否再lookUp中，若在则说明是相交的节点，否则查询完毕不存在相交点

code1:hash map

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

class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
        #简单思路：哈希表法
        lookUp = set()#存储A的所有节点
        #先存储A所有节点
        curA = headA
        while curA:
            lookUp.add(curA)
            curA = curA.next
        #再判断B的节点是否再lookUp中，若在则说明是相交的节点
        cur_B = headB
        while cur_B:
            if cur_B in lookUp: return cur_B
            cur_B = cur_B.next
        else: return None#查询完不存在相交点

解法2：双指针

第一次遍历找到长度差，第二次长链先走长度差，再找交叉节点

code2: bi-pointer

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
        #简单思路：双指针（两次遍历），第一次遍历找到长度差，第二次长链先走长度差，再找交叉节点
        #first: find the length difference 
        curA,curB = headA,headB
        numA = numB = 0
        while curA:
            numA += 1
            curA = curA.next
        while curB:
            numB += 1
            curB = curB.next
        diff = abs(numA-numB)
        #second: let the long chain go `diff` first, then to get the intersection.
        curA,curB = headA,headB
        if numA<numB:#let long chain go `diff` steps first
            for i in range(diff):
                curB = curB.next
        else:
            for i in range(diff):
                curA = curA.next
        while curA:#get the intersection
            if curA == curB: return curA
            curA = curA.next
            curB = curB.next
        return None