160. Intersection of Two Linked Lists (List；Two-Pointers)

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：如何让两个指针同步到达交叉点。当链表A访问到结尾后付给链表B的头指针，同理对链表B操作，从而在第二遍遍历时必定是同步的。

/**
 * 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* curA = headA;
        ListNode* curB = headB;
        
        while(curA && curB){
            if(curA == curB){
                return curA;
            }
            
            curA = curA->next;
            curB = curB->next;
            
        }
        if(!curA && !curB) return NULL;
        else if(!curA){
            curA = headB;
            while(curB){
                curB = curB->next;
                curA = curA->next;
            }
            curB = headA;
        }
        else{
            curB = headA;
            while(curA){
                curB = curB->next;
                curA = curA->next;
            }
            curA = headB;
        }
        
        while(curA){
            if(curA == curB){
                return curA;
            }
            
            curA = curA->next;
            curB = curB->next;
        }
        return NULL;
    }
};