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[leetcode]Linked List Cycle II

2013-12-03 21:32 531 查看
http://oj.leetcode.com/problems/linked-list-cycle-ii/

老题。当快慢指针相交时,通过方程或观察可知,从head到环开始点的距离和从相遇点开始是一样的,那么从相遇点和从开始点再一起走直到相遇就行了。

/**
* 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) {
ListNode *fast = head;
ListNode *slow = head;
do {
if (fast == NULL || fast->next == NULL) return NULL;
fast = fast->next->next;
slow = slow->next;
} while (fast != slow);
ListNode *n1 = head;
ListNode *n2 = slow;
while (n1 != n2) {
n1 = n1->next;
n2 = n2->next;
}
return n1;
}
};
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