LeetCode题解160-Intersection of Two Linked Lists(C++)

题目：

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：设定p1、p2两个指针，分别从两个链表头端往尾端依次走。当一个指针走到了尾节点，则令其从另一个链表头端再次出发，另一个指针也是如此。在第二次途中必定相遇。A链表头节点到交点的距离为D1，B链表头节点到交点距离为D2，交点到尾端的距离为D3。因为两个指针的总距离皆为D1+D2+D3。(如果第一次就相遇的话，D1 == D2)

代码：

/**
 * 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* ptr1 = headA;
        ListNode* ptr2 = headB;
        if(ptr1 == NULL || ptr2 == NULL) return NULL;
        while(ptr1 !=ptr2){
            ptr1 = ptr1->next;
            ptr2 = ptr2->next;
            if(ptr1 == ptr2) break;
            if(ptr1 == NULL) ptr1 = headB;
            if(ptr2 == NULL) ptr2 = headA;
        }
        return ptr1;
    }
};