160. Intersection of Two Linked Lists

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#链表交叉节点 #LeetCode #160 #Intersect

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本文介绍了一种在两个单链表中寻找它们开始相交节点的方法。通过将其中一个链表的末尾连接到另一个链表的头部形成环，并利用快慢指针技巧找到环的入口节点，以此确定两链表的交点。

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.

分析：
复用Linked List Cycle II的思路，找到链表的尾部节点，连上A的头节点，构成一个环，找到环的入口节点即可。
注意：返回入口节点前，记得将环打开，复原原有的结构。

/**
 * 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;

        ListNode *cur = headA;
        while (cur->next) {
            cur = cur->next;
        }

        cur->next = headA;

        ListNode *result = detectCycle(headB);
        cur->next = NULL;

        return result;
    }

    ListNode *detectCycle(ListNode *head) {
        ListNode *fast = head, *slow = head;
        while (fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;

            if (slow == fast) {
                ListNode *slow2 = head;
                while (slow != slow2) {
                    slow = slow->next;
                    slow2 = slow2->next;
                }
                return slow2;
            }
        }
        return NULL;
    }
};