LeetCode : 160. 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。提供了两种思路：一种利用栈实现，另一种通过计算链表长度并调整遍历起点来达到同步遍历，最终找到交点。该算法力求实现O(n)的时间复杂度和O(1)的空间复杂度。

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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.

分析：

思路1：利用栈后进先出的特性，将链表元素放入栈，然后从尾开始比较，会用O（n）的空间复杂度

思路2：先求出两个链表的长度lengthA、lengthB，然后长的链表先后移使之与短的长度相同，然后开始便利，找到相同的节点。

AC代码如下：

 ListNode *getIntersectionNode(ListNode *headA, ListNode *headB)
    {
        ListNode * res = NULL;
        int lengthA = 0;
        int lengthB = 0;
        ListNode *a,*b;
        a = headA;
        b = headB;
        while(a != NULL)
        {
            lengthA++;
            a = a->next;
        }
        while(b != NULL)
        {
            lengthB++;
            b = b->next;
        }
        a = headA;
        b = headB;
        int diff = abs(lengthA - lengthB);
        if(lengthA > lengthB)
        {
            for(int i = 0;i < diff;i++)
                a = a->next;
        }
        if(lengthA < lengthB)
        {
            for(int i = 0;i < diff;i++)
                b = b->next;
        }
        while(a!=NULL && b!= NULL)
        {
            if(a == b)
            {
                res = a;
                break;
            }
            a = a->next;
            b = b->next;
        }
        return res;
        
    }