[160] Intersection of Two Linked Lists

總是想要先寫出來再說......
好像不是一個好習慣 ？！給兩個linked list
找出它們從哪裡開始重疊
（或者說從哪裡開始是一樣的)

我好笨解法:
Intersection of Two Linked Lists
/**
 * 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;
    if (headA == headB)
        return headA;
    struct ListNode *t1, *t2;
    for(t1=headA;t1->next != NULL; t1 =t1->next)
    {
        for(t2=headB;t2->next != NULL; t2 =t2->next)
        {
            if (t1 == t2)
                return t1;
        }
    }

    for(t2=headB;t2->next != NULL; t2 =t2->next)
    {
        if (t1 == t2)
            return t2;
    }
    if (t1 == t2)
        return t2;
   
    return NULL;
}

然後就是看了別人解法果然還是我真的好笨.......
/**
 * 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;
    if (headA == headB)
        return headA;
    struct ListNode *t1, *t2;
    int i=0,l1=1,l2=1;
    for(t1=headA;t1->next != NULL; t1 =t1->next)
        l1++;

    for(t2=headB;t2->next != NULL; t2 =t2->next)
        l2++;

    t1 = headA;
    t2 = headB;      

    if (l1 > l2)
        for(i=0;i++<(l1-l2); t1 =t1->next)
            ;
    else if (l1 < l2)
        for(i=0;i++<(l2-l1); t2 =t2->next)
            ;
   
    while(t1 != NULL && t2 != NULL)
    {
        if (t1 == t2)
            return t1;
        t1 = t1->next;
        t2 = t2->next;
    }
       
    return NULL;
}

(怎麼沒辦法插入繼續閱讀 XD）

[20251110] 更新

原來當年之後已經有好精簡寫法了Orz 偷看了答案再寫，寫出了無窮迴圈XD

要注意原本的headA跟headB在後面還會用到，所以不能讓它自己跑去 next，

要另外用pointer來寫～以上。

struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {

struct ListNode *p1 = headA;

struct ListNode *p2 = headB;

while (p1 != p2)

{

if (p1==NULL)

p1 = headB;

else

p1 = p1->next;

if (p2==NULL)

p2 = headA;

else

p2 = p2->next;

}

return p1;

}