02.07

解法一：如果两个链表有相交，那么从后面看一定是相同的，所以只需要把长的移动到和短的链表一样的长度开始遍历即可

class Solution {
public:
    ListNode* getIntersectionNode(ListNode* headA, ListNode* headB) {
        ListNode* cur1 = headA;
        ListNode* cur2 = headB;
        int l1 = 0, l2 = 0;

        // 求出两个链表长度
        while (cur1 != NULL) {                  
            l1++;
            cur1 = cur1->next;
        }
        while (cur2 != NULL) {
            l2++;
            cur2 = cur2->next;
        }

        // 此时 cur1 cur2 都指向空，所以要把他们指向 head
        cur1 = headA;
        cur2 = headB;

        // 把长的设置成1
        if (l1 < l2) {        
            swap(l1, l2);
            swap(cur1, cur2);
        }

        // 把长的移动到和短的一样的位置
        int l = l1 - l2;
        while (l--) {
            cur1 = cur1->next;
        }

        // 从相等处开始遍历
        while (cur1 != NULL) {
            if (cur1 == cur2) {
                return cur1;
            }
            cur1 = cur1->next;
            cur2 = cur2->next;
        }

        return NULL;
    }
};

解法二：太强了，暂时理解不了

class Solution {
public:
    ListNode* getIntersectionNode(ListNode* headA, ListNode* headB) {
        ListNode* cur1 = headA;
        ListNode* cur2 = headB;

        while (cur1 != cur2) {
            cur1 = cur1 != nullptr ? cur1->next : headB;
            cur2 = cur2 != nullptr ? cur2->next : headA;
        }
        return cur1;
    }
};