https://leetcode.com/problems/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. Credits: Special thanks to @stellari for adding this problem and creating all test cases.
问题思路：
先判断是否相交，如何判断呢，如果俩个链表相交则必然最后一个节点的值是相等的。如果不相等则表示不相交。
如果相交则返回相交节点。考虑一个这样的事实，如果相交的话，链表的最后一段是相同的，那么我们可以先遍历俩个链表的长度，长的那个链表先遍历多出来的那部分，然后同时遍历两个链表，并且一边遍历一边比较，直到遍历到第一个相同的节点，那么这个节点就是第一个相交的节点。
代码如下/** * 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) {// 基础判断if(headA == NULL || headB == NULL){return NULL;}// 判断是否有相交，若无相交则返回NULLListNode* pA = headA, *pB = headB;int countA = 0, countB = 0;while(pA->next != NULL) {pA = pA->next;++countA;}while(pB->next != NULL) {pB = pB->next;++countB;}if (pA->val != pB->val) {return NULL;}// 有相交,返回相交节点pA = headA;pB = headB;int dif = countA – countB;if (dif > 0) {while (dif–) pA = pA->next;}else{while (dif++) pB = pB->next;}while (pA->val != pB->val){pA = pA->next;pB = pB->next;}return pA;}};
不敢接受失败的人，往往是那些追求完美的人，
