c_cpp 160.两个链接列表的交集 - 简单 - 2018.8.6

Posted 2021-05-14

/**
1：把 LA 的尾巴链接到 LA 的头部，创造一个循环
2：按照题目 142 的解法来找到连接点
 */
 
class Solution {
public:
    ListNode *detectCycle( ListNode *head ) {
        ListNode *slow = head, *fast = head;
        while (fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
            if (slow == fast) {
                ListNode *slow2 = head;
                while (slow2 != slow) {
                    slow2 = slow2->next;
                    slow = slow->next;
                }
                return slow2;
            }
        }
        return NULL;
    }
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if (headA == NULL || headB == NULL) return NULL;
        
        // 先制造一个 loop
        ListNode *tailNodeA = headA;
        while (tailNodeA->next != NULL) {
            tailNodeA = tailNodeA->next;
        }
        tailNodeA->next = headA;
                
        // 找链表相遇的节点
        ListNode *meetNode = detectCycle(headB);

        // 恢复现场
        tailNodeA->next = NULL;
        
        return meetNode;
    }
};