160. 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.

思路：

找链表交界，很类似Linked List Cycle II那题，方法也是类似的双指针相遇法。分两步走：

1. 如何判断两链表是否相交？

两链表相交则他们必然有共同的尾节点。所以遍历两个链表，找到各自的尾节点，如果tailA!=tailB则一定不相交，反之则相交。

2. 如何判断两链表相交的起始节点？

在第1步判断相交时可以顺带计算两链表的长度lenA和lenB。让长的链表的head先走abs(lenA-lenB)步，然后和短链表的head一起走，直到两者相遇，即为要找的节点。

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if(!headA || !headB) return NULL;
        int lenA = 0, lenB = 0;
        ListNode *tailA = headA, *tailB = headB;
        
        while(tailA->next) {
            tailA = tailA->next;
            lenA++;
        }
        while(tailB->next) {
            tailB = tailB->next;
            lenB++;
        }
        if(tailA!=tailB) return NULL;
        
        if(lenA>lenB) {
            for(int i=0; i<lenA-lenB; i++)
                headA = headA->next;
        }
        else {
            for(int i=0; i<lenB-lenA; i++)
                headB = headB->next;
        }
        
        while(headA!=headB) {
            headA = headA->next;
            headB = headB->next;
        }
        
        return headA;
    }
};