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Linked List Cycle II
Total Accepted: 20444 Total Submissions: 66195My Submissions
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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?
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题意：给定一个单链表，判断该链表中是否存在环，如果存在，返回环开始的节点
思路：
1.定义两个指针，快指针fast每次走两步，慢指针s每次走一次，如果它们在非尾结点处相遇，则说明存在环
2.若存在环，设环的周长为r，相遇时，慢指针走了 slow步，快指针走了 2s步，快指针在环内已经走了 n环，
则有等式 2s = s + nr 既而有 s = nr
3.在相遇的时候，另设一个每次走一步的慢指针slow2从链表开头往前走。因为 s = nr，所以两个慢指针会在环的开始点相遇
复杂度：时间O(n)
struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
ListNode *detectCycle(ListNode *head) {
if(!head || !head->next) return NULL;
ListNode *fast, *slow, *slow2;
fast = slow = slow2 = head;
while(fast && fast->next){
fast = fast->next->next;
slow = slow->next;
if(fast == slow && fast != NULL){
while(slow->next){
if(slow == slow2){
return slow;
}
slow = slow->next;
slow2 = slow2->next;
}
}
}
return NULL;
}
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