letecode [160] - Intersection of Two Linked Lists

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

题目大意：

　　给定两个链表，找到它们的第一个公共节点并返回。

理解：

　　从第一个公共节点开始，两链表后面的节点都是公共的。

　　计算两个链表的长度，从长度相同的节点开始往后遍历，并比较p==q，若相等，则当前节点为第一个公共节点；不等则继续往后找。

代码 C++：

/*** 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;int lenA=0,lenB=0;ListNode *p = headA;ListNode *q = headB;while(p!=NULL){lenA++;p = p->next;}while(q!=NULL){lenB++;q = q->next;}p = headA;q = headB;while(lenA>lenB){p = p->next;lenA--;}while(lenA<lenB){q = q->next;lenB--;}while(p!=NULL){if(p==q){return p;}p = p->next;q = q->next;}return NULL;}
};

运行结果：

　　执行用时 :72 ms, 在所有C++提交中击败了84.85%的用户

内存消耗 :16.9 MB, 在所有C++提交中击败了11.06%的用户

转载于:https://www.cnblogs.com/lpomeloz/p/11002364.html

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