题解 | #链表中环的入口结点#

快慢指针法判断是否有环，若有环，则快慢指针相遇结点与头结点依次后移一位会相遇在入口结点

时间复杂度O(n)，空间复杂度：O(1)

struct ListNode* EntryNodeOfLoop(struct ListNode* pHead ) {
    // write code here
    struct ListNode* head = pHead;
    struct ListNode* fast = pHead;
    struct ListNode* slow = pHead;
    while (fast && fast->next) {
        fast = fast->next->next;
        slow = slow->next;
        if (fast == slow) {
            while (fast != head) {
                fast = fast->next;
                head = head->next;
            }
            return fast;
        }
    }
    return NULL;
}