160 相交链表

 

 

 

 

 

 

 

 

指针 pA 指向 A 链表，指针 pB 指向 B 链表，依次往后遍历
如果 pA 到了末尾，则 pA = headB 继续遍历
如果 pB 到了末尾，则 pB = headA 继续遍历
比较长的链表指针指向较短链表head时，长度差就消除了
如此，只需要将最短链表遍历两次即可找到位置

作者：reals
链接：https://leetcode-cn.com/problems/intersection-of-two-linked-lists/solution/tu-jie-xiang-jiao-lian-biao-by-user7208t/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
            if(headA==NULL || headB==NULL)
            {
                return NULL;
            }
            ListNode* a=headA;
            ListNode* b=headB;
            while(a!=b)
            {
                a= a==NULL?headB:a->next;
                b= b==NULL?headA:b->next;
            }
            return a;

    }
};

 

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标签：ListNode,相交,next,链表,headB,headA,NULL,160
来源： https://www.cnblogs.com/libin123/p/15314710.html