Leetcode 141. Linked List Cycle

最新推荐文章于 2025-05-07 22:36:04 发布

原创最新推荐文章于 2025-05-07 22:36:04 发布 · 276 阅读

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本文介绍了一种使用快慢指针的方法来检测链表是否存在循环，并进一步确定循环的入口位置。此外，还讨论了如何将这种方法应用于查找数组中的重复数字问题。

problems:
https://leetcode.com/problems/linked-list-cycle/description/
https://leetcode.com/problems/linked-list-cycle-ii/description/
https://leetcode.com/problems/find-the-duplicate-number/description/
discuss:
https://discuss.leetcode.com/topic/2975/o-n-solution-by-using-two-pointers-without-change-anything
https://discuss.leetcode.com/topic/25580/two-solutions-with-explanation-o-nlog-n-and-o-n-time-o-1-space-without-changing-the-input-array
reference:
http://keithschwarz.com/interesting/code/?dir=find-duplicate

Given a linked list, determine if it has a cycle in it.

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

碰到很有意思的一组题目，记录一下。

不限定空间的话，那么O(n)记忆一下，就解出来了。
但是要O(1)的空间的话，就没什么头绪。
翻了下discuss，发现这题很有意思。

用快慢指针的想法，假如存在cycle，那么必定在某点相遇

// initial-state: fast = slow = head
while (fast->next != nullptr && fast->next->next != nullptr) {
    slow = slow->next;
    fast = fast->next;
    if (slow == fast) return true;
}
return false;

进一步，怎么确定cycle的入口entry呢？
假设相遇在k处，入口在s处，相遇点距离入口m处，cycle长度为r，
那么slow走了k，fast走了2k

  |<-          k           ->|
  |<-   s    ->|<-    m    ->|
head ------- entry --->---meeting--->---
               ^                       |
               |                       v
               ------------<------------

k + nr = 2k
=> k = nr
=> k = (n-1)r + r
=> k-m = (n-1)r + r-m
=> s = (n-1)r + r-m
=> s = r-m

意味着相遇点在走r-m=s步就回到了入口处

// initial-state: fast = slow = entry = head
while (fast->next != nullptr && fast->next->next != nullptr) {
    slow = slow->next;
    fast = fast->next;
    if (slow == fast) {
        while (entry != slow) {
            entry = entry->next;
            slow = slow->next;
        }
        return entry;
    }
}
return nullptr;

另外在看看这题 287. Find the Duplicate Number