160. 相交链表 – Blog of Aspirer

题目链接：https://leetcode.cn/problems/intersection-of-two-linked-lists/

下面两种方法都想到了，官方题解的双指针法想不到。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
/* 暴力解法
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        ListNode* hb = headB;
        while (headA) {
            headB = hb;
            while (headB) {
                if (headB == headA) {
                    return headA;
                }
                headB = headB->next;
            }
            headA = headA->next;
        }
        return NULL;
    }
*/

    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        unordered_set<ListNode*> s;
        while (headA) {
            s.insert(headA);
            headA = headA->next;
        }
        while (headB) {
            if (s.count(headB) > 0) {
                return headB;
            }
            headB = headB->next;
        }
        return NULL;
    }
};

/**

* Definition for singly-linked list.

* struct ListNode {

* int val;

* ListNode *next;

* ListNode(int x) : val(x), next(NULL) {}

* };

class Solution {

public:

/* 暴力解法

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

ListNode* hb = headB;

while (headA) {

headB = hb;

while (headB) {

if (headB == headA) {

return headA;

}

headB = headB->next;

}

headA = headA->next;

}

return NULL;

}

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

unordered_set<ListNode*> s;

while (headA) {

s.insert(headA);

headA = headA->next;

}

while (headB) {

if (s.count(headB) > 0) {

return headB;

}

headB = headB->next;

}

return NULL;

}

};