【LeetCode】142 - Linked List Cycle II

标签：

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?

Solution：

Discuss上的分析：Suppose the first meet at step k,the length of the Cycle is r. so..2k-k=nr,k=nrNow, the distance between the start node of list and the start node of cycle is s. the distance between the start of list and the first meeting node is k(the pointer which wake one step at a time waked k steps).the distance between the start node of cycle and the first meeting node ism, so...s=k-m, s=nr-m=(n-1)r+(r-m),here we takes n = 1..so, using one pointer start from the start node of list, another pointer start from the first meeting node, all of them wake one step at a time, the first time they meeting each other is the start of the cycle.

/**
 * 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 || !head->next)return NULL;
        ListNode *faster = head;
        ListNode *slower = head;
        while(faster->next && faster->next->next){
            faster=faster->next->next;
            slower=slower->next;
            if(faster==slower){
                ListNode *temp=head;
                while(temp!=slower){
                    temp=temp->next;
                    slower=slower->next;
                }
                return temp;
            }
        }
        return NULL;
    }
};

【LeetCode】142 - Linked List Cycle II

标签：

原文地址：http://www.cnblogs.com/irun/p/4739753.html