[Leetcode]-Intersection of Two Linked Lists

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本博客介绍了一个算法，用于找出两个单链表中相交的节点。该算法通过计算链表长度并使较长的链表先行移动至相交点前，然后同时遍历两个链表直到找到相交节点。

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.

题目：找出两个单链表中的交叉点
注意：在交叉节点之后的所有节点都是相同的节点，包括长度也是一样的。
1、方法就是首先分别求出两个单链表的长lA和lB
2、长的链表提前向前走| lA-lB |步
3、两个链表同步向前走，每走一步判断是否是交叉点，知道链表尾端

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
    if(headA == NULL  || headB == NULL)
        return NULL;
    struct ListNode *PA ;
    struct ListNode *PB ;
    PA = headA;
    PB = headB;
    int lA = 0 ;
    int lB = 0 ;
    while(headA)
    {
        lA++;
        headA = headA->next;
    }
    while(headB)
    {
        lB++;
        headB = headB->next;
    }
    int i = 0;
    int j = 0;

    if(lA > lB)
    {
        for(i=0;i<lA-lB;i++)
        {
            PA = PA->next;
        }
    }
    else
    {
        for(i=0;i<lB-lA;i++)
        {
            PB = PB->next;
        } 
    }

    while(PA != NULL && PB != NULL)
    {
        if(PA == PB)
            return PA;
        PA = PA->next;
        PB = PB->next;
    }

    return NULL;   
}