LeetCode题目：Linked List Cycle II

来源：互联网发布：SQL中怎么定义联合主键编辑：程序博客网时间：2024/06/18 11:55

这个题目，首先应当找出环内任意一点：俩指针，一快一慢（走一步，走两步），重合就是环内节点。

然后将问题转化为经典问题：寻找链表交叉节点位置。

P.S. 创建了一个GitHub项目，目前已经更新了大概十道题：https://github.com/xlinc/CrackingLeetCode.git

class Solution {public:ListNode *detectCycle(ListNode *head) {if (head == NULL)return NULL;ListNode *p1 = head, *p2 = head->next;if ( p2 == NULL)return NULL;if (p2 == head){return head;}while (p1 != p2){p1 = p1->next;if (p2->next != NULL){if (p2->next == head)return head;p2 = p2->next;if (p2->next != NULL){if (p2->next == head)return head;p2 = p2->next;}elsereturn NULL;}elsereturn NULL;}p1 = head;ListNode* innerNode = p2;//cout <<"innerNode: "<< innerNode->val << endl;ListNode* checkNode = innerNode->next;int len1 = 0, len2 = 0;do{len1++;p2 = p2->next;} while (p2 != innerNode);//cout << len1 << endl;do{len2++;p1 = p1->next;} while (p1 != innerNode);p1 = (len1 > len2 ? innerNode : head);p2 = (len1 > len2 ? head : innerNode);for (int i = 0; i < abs(len1-len2); i++){p1 = p1->next;}while (p1!= p2){p1 = p1->next;p2 = p2->next;}return p1;}};

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