Intersection of Two Linked Lists

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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。通过首先确定两个链表的长度，再同步遍历两个链表直至找到交点。此算法运行时间为O(n)，仅使用O(1)额外空间。

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一、问题描述

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.

二、思路

先确定AB两个链表的大小，然后找到两个链表的交点。

三、代码

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if(headA == NULL || headA == NULL) 
            return nullptr;
        ListNode *currentA = headA;
        int sizeA = 0;
        while(currentA){
            ++sizeA;
            currentA = currentA -> next;
        }
        ListNode *currentB = headB;
        int sizeB = 0;
        while(currentB){
            ++sizeB;
            currentB = currentB -> next;
        }
        currentA = headA;
		currentB = headB;
		while(sizeA > sizeB){
		    --sizeA;
		    currentA = currentA -> next;
		}
        while(sizeB > sizeA){
		    --sizeB;
		    currentB = currentB -> next;
		}
		while(currentA != currentB) {
			currentA = currentA->next;
			currentB = currentB->next;
		}
		return currentA;
    }
};