面试笔试杂项积累-leetcode 156-160

160.160-Intersection of Two Linked Lists-Difficulty: Easy

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

A:          a1 → a2
↘
c1 → c2 → c3
↗
B:     b1 → b2 → b3

begin to intersect at node c1.

Notes:

If the two linked lists have no intersection at all, return

null

.
The linked lists must retain their original structure after the function returns.

You may assume there are no cycles anywhere in the entire linked structure.
Your code should preferably run in O(n) time and use only O(1) memory.
Credits:

Special thanks to
@stellari for adding this problem and creating all test cases.

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思路

两个链表指向一个链表，找到汇聚的起点

解题思路:1.首先想到的是哈希，先将一个链表中结点都插入字典中，然后遍历另一个链表中的结点看是否有结点在字典中;但这种方法需要开辟较大的内存空间来存储字典；

2.双指针法.首先对其中一个链表头尾连接.那么问题就变成了看另一个链表是否存在环的问题了.但这种方法改变了原本链表的结构，需要在最后对其还原；

3.先计算两个链表的长度差，然后对长链表头结点开始移动长度差长的结点,找到位置对应的结点.然后逐个比较是否相等;

第二种方法就和之前的一道题一样，那就没有意义了

博主选择了最机智的第三种方法

/**
* Definition for singly-linked list.
* public class ListNode {
* public int val;
* public ListNode next;
* public ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode GetIntersectionNode(ListNode headA, ListNode headB) {
ListNode A_temp = headA;
ListNode B_temp = headB;
int A_Length = 0;
int B_Length = 0;
while (A_temp != null || B_temp != null)
{
if (A_temp != null)
{
++A_Length;
A_temp = A_temp.next;
}
if (B_temp != null)
{
++B_Length;
B_temp = B_temp.next;
}
}

A_temp = headA;
B_temp = headB;
if (A_Length > B_Length)
{
for (int i = 0; i < A_Length - B_Length; i++)
A_temp = A_temp.next;
}
else if (B_Length > A_Length)
{
for (int i = 0; i < B_Length - A_Length; i++)
B_temp = B_temp.next;
}

while (A_temp != null)
{
if (A_temp == B_temp)
return A_temp;
A_temp = A_temp.next;
B_temp = B_temp.next;
}
return null;
}
}