53-Linked List Cycle II

1. Linked List Cycle II My Submissions QuestionEditorial Solution
   Total Accepted: 74093 Total Submissions: 235430 Difficulty: Medium
   Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

Note: Do not modify the linked list.

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

思路：假设快慢指针在x处相遇，
此时快指针在环内假设走了n圈加x步，而总的步数：nr+x+a
是慢指针的2倍
即有nr+x+a =(x+a)*2
上式为a=(n-1)r+r-x;
说明慢指针走到a时候，快指针从x处走了r-x前一个加上n-1圈的距离，
注意此时快指针从X处一步一步走，慢指针也一步一步走，直到在入口相遇

时间复杂度： O(n)
空间复杂度： O(1)

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        if(head==NULL||head->next==NULL)return NULL;
        ListNode *p_quick=head,*p_slow=head;
        while(p_quick!=NULL&&p_quick->next!=NULL){
            p_quick = p_quick->next->next;
            p_slow = p_slow->next;
            if(p_quick==p_slow)break;
        }
        if(p_quick==p_slow){
            p_slow=head;
            while(p_slow!=p_quick){
                p_slow = p_slow->next;
                p_quick=p_quick->next;
            }
            return p_slow;
        }
        else return NULL;
    }
};