Intersection of Two Linked Lists—

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.

开始想道用栈来做，先入栈，然后出栈比较直接就出结果了，但是题目说只能利用一个空间，所以栈的方法不行了。

看网上的答案，没想到这么简单，暴力解法，什么算法、数据结构都没用，题刷多了，脑袋都僵了，一直以为要用什么算法来做呢。。。。

思路：

查找两个链表的第一个公共节点，如果两个节点的尾节点相同，肯定存在公共节点

方法：长的链表开始多走（h1的数量 - h2的数量）步，然后和短链表同步往下走，遇到的第一个相同的节点就是最早的公共节点

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode(int x) : val(x), next(NULL) {}* };*/
class Solution {
public:ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {if(headA==NULL||headB==NULL)return NULL; ListNode *flagA=headA;ListNode *flagB=headB;int countA=1;int countB=1;while(flagA->next!=NULL){countA++;flagA=flagA->next;}while(flagB->next!=NULL){countB++;flagB=flagB->next;}if(flagA!=flagB)return NULL;else{int diff=abs(countA- countB);if(countA>=countB){flagA=headA;flagB=headB;}else{flagA=headB;flagB=headA;}while(diff){flagA=flagA->next;diff--;}while(flagA!=flagB){flagA=flagA->next;flagB=flagB->next;}return flagA;}}
};

转载于:https://www.cnblogs.com/qiaozhoulin/p/4758718.html

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