Intersection of Two Linked Lists

寻找两个单链表相交节点的算法实现

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本文提供了一种高效的算法来查找两个单链表相交的起始节点，通过比较链表长度并调整指针位置，最终实现O(n)时间复杂度和O(1)内存使用。

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.

class Solution {
public:
	ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) 
	{
		if (0 == headA || 0 == headB) return 0;
		int lenA = 0, lenB = 0;
		ListNode A(0), B(0), *pA = &A, *pB = &B;
		pA->next = headA;
		pB->next = headB;

		while(pA->next != 0)
		{
			++lenA;
			pA = pA->next;
		}
		while(pB->next != 0)
		{
			++lenB;
			pB = pB->next;
		}
		if (pA != pB) 
			return 0;
		int diff = abs(lenA - lenB);
		pA = &A; // make a fatal endless loop mistak here.
		pB = &B;
		pA->next = headA;
		pB->next = headB;
		if (lenA > lenB)
		{
			while (diff-- > 0)
				pA = pA->next;
		}
		else if (lenA < lenB)
		{
			while (diff-- > 0)
				pB = pB->next;
		}
		while (1)
		{
			if (pA->next == pB->next)
				return pA->next;
			pA = pA->next;
			pB = pB->next;
		}        	
	}
};