您的位置:首页 > 职场人生

Leetcode Linked List Cycle II

2013-11-12 19:58 417 查看
解法参考:http://blogread.cn/it/article/2452?f=hot1

先判断有没有环,有环的话,设两个指针,一个指针在head,另一个指针在相遇的点,两个指针同时往前走,则这两个指针相遇的点就是环的开始节点。

/**
* 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 *cross = findCross(head);
ListNode *tmp = head;
if(cross != NULL){
while(tmp != cross){
cross = cross->next;
tmp = tmp->next;
}
}
return cross;
}
ListNode *findCross(ListNode *head){
if(head == NULL || head->next == NULL)    return NULL;
ListNode *cross = NULL, *one = head->next, *two = head->next->next;
while(one != NULL && two != NULL){
if(one == two){
cross = one;
return cross;
}
if(two->next == NULL)    return cross;
one = one->next;
two = two->next->next;
}
return cross;
}
};
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签:  OJ leetcode 面试