LeetCode_Lined List CycleII

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?

这道题让我很无语，感觉我的算法和AC的应该是一样的，但是就是TLE，贴出来：

/*** 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) {// IMPORTANT: Please reset any member data you declared, as// the same Solution instance will be reused for each test case.ListNode* slow = head;ListNode* fast = head;do{if( !slow || !fast ) return NULL;slow = slow->next;fast = fast->next;if( fast ) fast = fast->next;else return NULL;}while( slow != fast );slow = head;while( slow != fast ){slow = slow->next;fast = fast->next;}return slow;}
};

上面是AC的，下面是自己写的TLE：

/*** 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){return NULL;}ListNode *pSingleSpeed=head;ListNode *pDpubleSpeed=head;while (pSingleSpeed!=NULL&&pDpubleSpeed!=NULL){pSingleSpeed=pSingleSpeed->next;pDpubleSpeed=pDpubleSpeed->next;if (pDpubleSpeed!=NULL){pDpubleSpeed=pDpubleSpeed->next;}/*  else{return NULL;}*/if (pSingleSpeed==pDpubleSpeed/*&&pSingleSpeed!=NULL&&pDpubleSpeed!=NULL*/){break;}}if (pDpubleSpeed==NULL||pSingleSpeed==NULL){return NULL;}pDpubleSpeed=head;while (pDpubleSpeed!=pSingleSpeed){pSingleSpeed=pDpubleSpeed->next;pSingleSpeed=pSingleSpeed->next;}return pSingleSpeed;}
};

两者思想都是一样的，先判断是否具有回路，如果有，让一个指针回到起点，两个指针以相同速度继续走，在此相遇地点就是环的入口，如果不清楚，画图一看就知道了。

LeetCode_Lined List CycleII

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