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.

方法：判断每条路径的长度，利用两条路径之差(step)，然后在长的那条路径上提前走step步。

/** * 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) {        int lenA = 0,lenB = 0;        ListNode *cur = headA;        while(cur!=NULL){            lenA++;            cur=cur->next;        }        cur = headB;        while(cur!=NULL){            lenB++;            cur=cur->next;        }        ListNode *another;        int step = abs(lenA-lenB);        if(lenA>lenB){            cur = headA;            another = headB;        }        else{            cur = headB;            another = headA;        }        while(step--){            cur=cur->next;        }        while(cur!=another){            cur = cur->next;            another = another->next;        }        return cur;    }};

方法2：简易实现

/** * 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) {       ListNode *cur1 = headA,*cur2 = headB;       while(cur1!=cur2){           cur1 = cur1 ? cur1->next:headB;           cur2 = cur2 ? cur2->next:headA;       }       return cur1;    }};