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Solution142.hpp
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71 lines (60 loc) · 2.52 KB
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//
// Solution142.hpp
// Algorithm
//
// Created by Pancf on 2019/12/21.
// Copyright © 2019 Pancf. All rights reserved.
//
#ifndef Solution142_hpp
#define Solution142_hpp
#include <stdio.h>
class Solution142 {
/**
Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.
Note: Do not modify the linked list.
Example 1:
Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the second node.
Example 2:
Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.
Example 3:
Input: head = [1], pos = -1
Output: no cycle
Explanation: There is no cycle in the linked list.
Follow-up:
Can you solve it without using extra space?
================================================================================================
Accept details:
Runtime: 12 ms, faster than 76.79% of C++ online submissions for Linked List Cycle II.
Memory Usage: 9.7 MB, less than 88.10% of C++ online submissions for Linked List Cycle II.
思路：此题是141题的扩展，不仅仅是需要判断是否有环，还需要找到环的入口（即第一个进入环的链表结点）。使用141中快慢指针的做法，
快、慢指针相遇时，相遇点在环内某个结点，快指针走的步数刚好是慢指针的两倍
start----cycle entry------------meeting-----------end
由此可得 length(start, meeting) = length(meeting, end) + length(cycle entry, meeting) => length(start, cycle entry) = length(meeting, end)
因此在快慢指针相遇时另增一个新指针从start开始，和慢指针一样每次loop向前走一步，两者相遇点必然在cycle entry
*/
public:
struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
ListNode *detectCycle(ListNode *head);
static void test() {
ListNode node1{1};
ListNode node2{2};
ListNode node3{3};
ListNode node4{4};
node1.next = &node2;
node2.next = &node3;
node3.next = &node4;
node4.next = &node2;
Solution142 s = Solution142();
printf("%d", s.detectCycle(&node1)->val);
}
};
#endif /* Solution142_hpp */