15. Linked List Cycle && Linked List Cycle II

Linked List Cycle

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

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

说明:两个指针不同步长。

/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
bool hasCycle(ListNode *head) {
ListNode *p1, *p2;
p1 = p2 = head;
while(p2 && p2->next && p2->next->next) {
p2 = p2->next->next;
p1 = p1->next;
if(p1 == p2) return true;
}
return false;
}
};

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?

说明:在上题基础上,将一个指针放到链表头,步长都设为1,相遇节点。(可以计算)

class Solution {
public:
ListNode *detectCycle(ListNode *head) {
ListNode *p1, *p2;
p1 = p2 = head;
while(p2 && p2->next && p2->next->next) {
p2 = p2->next->next;
p1 = p1->next;
if(p1 == p2) {
p1 = head;
while(p1 != p2) {
p1 = p1->next;
p2 = p2->next;
}
return p1;
}
}
return NULL;
}
};
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