leetcode刷题日记——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.

问题分析：题目目标是求二个链表是否相交，并且返回相交的节点。我们知道如果链表相交，则后面的点则全部相交，因此我们只需从二个链表的距离相交节点的相同距离处开始逐个节点比较。于是首先要求出二个链表分别长度，时间代价为m（假设m>n）,然后我们将较长的链表从m-n元素开始于较短的链表比较，这个过程最大代价我m,所以整个算法的最坏情况下代价为2m,满足O(n)的复杂度。实现代码如下：

/** * 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) {        if(headA==NULL||headB==NULL) return NULL;        int count1=0;int count2=0,k=0;        ListNode *p=headA,*q=headB;        while(p&&q){            p=p->next;            q=q->next;            count1++;count2++;        }        while(p){            p=p->next;            count1++;        }        while(q){            q=q->next;            count2++;        }        if(count2>=count1){            p=headA;            q=headB;            k=count2-count1;            while(k){                q=q->next;                k--;            }            while(p){                if(p==q) return p;                p=p->next;                q=q->next;            }            return NULL;                    }        else{            p=headA;            q=headB;            k=count1-count2;            while(k){                p=p->next;                k--;            }            while(p){                if(p==q) return p;                p=p->next;                q=q->next;            }            return NULL;        }            }};