【题解】【链表】【Leetcode】Linked List Cycle II

Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

Follow up:
Can you solve it without using extra space?

思路

这题是Linked List Cycle的进阶版

Given a linked list, determine if it has a cycle in it.

bool hasCycle(ListNode *head) {
    if(head == NULL) return false; //带环链表还要考虑只有单个元素的情况
    ListNode *faster = head, *slower = head;
    while(faster->next != NULL && faster->next->next != NULL && slower->next != NULL){//直接判断faster->next->next != NULL会抛错
        faster = faster->next->next;
        slower = slower->next;
        if(faster == slower) 
            return true;
    }
    return false;
}

faster和slower相遇之后必然在环上，让slower再走一圈计算环的长度len。另让两个指针p1,p2从head开始走，p1比p2先走len步，这样当p2走到环开始处时，正好p1与p2第一次相遇。

ListNode *detectCycle(ListNode *head) {
    if(head == NULL) return NULL;
    //带环链表要考虑只有单个元素的情况
    ListNode *faster = head, *slower = head;
    while(faster->next != NULL && faster->next->next != NULL && slower->next != NULL){//直接判断faster->next->next != NULL会抛错
        faster = faster->next->next;
        slower = slower->next;
        if(faster == slower){
            ListNode *p1 = head;//另让两个指针p1,p2从head开始走
            ListNode *p2 = head;
            slower = slower->next;//让slower再走一圈计算环的长度len
            p1 = p1->next;//设len是环的长度，p1比p2先走len步
            if(faster == slower) return faster;//自环
            while(faster != slower){
                slower = slower->next;
                p1 = p1->next;
            }
            while(p1 != p2){//当p1与p2第一次相遇时，正好p2走到环开始处
                p1 = p1->next;
                p2 = p2->next;
            }
            return p2;
        }
    }
    return NULL;
}