LeetCode刷题笔记(链表):linked-list-cycle-ii



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题目描述

Given a linked list, return the node where the cycle begins. If there is no cycle, returnnull.

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

解题思路

  • 同linked-list-cycle-i一题,使用快慢指针方法,判定是否存在环,并记录两指针相遇位置(Z);
  • 有环的情况下, 求链表的入环节点:遍历链表,把每个元素指向下个链表的指针赋值为NULL,则循环要么在链表结尾停止,要么在环状链表入口处停止。

C++版代码实现

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/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
//判断是否为环状链表
bool hasCycle(ListNode *head) {
if(head == NULL || head->next == NULL)
return false;
ListNode *fast = head;
ListNode *slow = head;
while(fast->next != NULL && fast->next->next != NULL){
slow = slow->next;
fast = fast->next->next;
if(slow->val == fast->val)
return true;
}
return false;
}

//检测函数
ListNode *detectCycle(ListNode *head) {
if(hasCycle(head)) {
ListNode *temp = NULL;
while(head ->next) {
temp = head ->next;
head ->next = NULL;
head = temp;
}
return head;
}
else
return NULL;
}
};

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完的汪(∪。∪)。。。zzz

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