160. Intersection of Two Linked Lists（C语言版本）

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本文介绍了一种在两个单链表中找到它们开始相交节点的方法。通过首先确定每个链表的长度，然后使较长链表向前移动以平衡两者长度，最后同步遍历并比较节点地址来找出交点。

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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.

解题思路：

1 分别求出两个单链表的长度；

2 将长度较长的那一个链表的节点向后移动，知道连个链表所处的长度相同；

3 依次比较此时链表的相同的位置是否指向相同的地址；

代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
    if(!headA || !headB) return NULL;
    int i=0,j=0;
    struct ListNode *pa=headA,*pb=headB;
    while(pa)
    {
        pa = pa->next;
        i++;
    }
    
    while(pb)
    {
        pb = pb->next;
        j++;
    }
    pa = headA;
    pb = headB;
    while(j<i)
    {
        pa = pa->next;
        i--;
    }
    while(i<j)
    {
        pb = pb->next;
        j--;
    }
    
    while(pa && pb)
    {
        if(pa == pb) 
            break;
        else
        {
            pa = pa->next;
            pb = pb->next;
        }
        
    }
    return pa;
    
    
}