floyd-cycle-finding | 易学教程

Linked list loop detection algorithm

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问题 I read some interview question online about how would you find if there's a loop in a linked list, and solution (Floyd's cycle-finding algorithm) is to have two pointers, one is 2x faster than the other, and check if they meet again. My question is: Why can't I just keep one pointer fixed, just move the other pointer forward by 1 step each time? 回答1: Because the first (non-moving) pointer might not lie within the loop, so the pointers would never meet. (Remember that a loop may consist of

Why does Floyd's cycle finding algorithm fail for certain pointer increment speeds?

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问题 Consider the following linked list: 1->2->3->4->5->6->7->8->9->4->...->9->4..... The above list has a loop as follows: [4->5->6->7->8->9->4] Drawing the linked list on a whiteboard, I tried manually solving it for different pointer steps, to see how the pointers move around - (slow_pointer_increment, fast_pointer_increment) So, the pointers for different cases are as follows: (1,2), (2,3), (1,3) The first two pairs of increments - (1,2) and (2,3) worked fine, but when I use the pair (1,3),

Floyd's cycle-finding algorithm

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问题 I'm trying to find this algorithm on C++ in .NET but can't, I found this one: // Best solution function boolean hasLoop(Node startNode){ Node slowNode = Node fastNode1 = Node fastNode2 = startNode; while (slowNode && fastNode1 = fastNode2.next() && fastNode2 = fastNode1.next()){ if (slowNode == fastNode1 || slowNode == fastNode2) return true; slowNode = slowNode.next(); } return false; } but doesn't seem to be right, or am I wrong? How can I actually prove that my hare will meet tortoise at

Floyd's cycle-finding algorithm

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I'm trying to find this algorithm on C++ in .NET but can't, I found this one: // Best solution function boolean hasLoop(Node startNode){ Node slowNode = Node fastNode1 = Node fastNode2 = startNode; while (slowNode && fastNode1 = fastNode2.next() && fastNode2 = fastNode1.next()){ if (slowNode == fastNode1 || slowNode == fastNode2) return true; slowNode = slowNode.next(); } return false; } but doesn't seem to be right, or am I wrong? How can I actually prove that my hare will meet tortoise at the end? Thanks in advance for any explanation how exactly does it work and proof EDITED About this

Detect period of unknown source

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问题 How to detect repeating digits in an infinite sequence? I tried Floyd & Brent detection algorithm but come to nothing... I have a generator that yields numbers ranging from 0 to 9 (inclusive) and I have to recognize a period in it. Example test case: import itertools # of course this is a fake one just to offer an example def source(): return itertools.cycle((1, 0, 1, 4, 8, 2, 1, 3, 3, 1)) >>> gen = source() >>> period(gen) (1, 0, 1, 4, 8, 2, 1, 3, 3, 1) 回答1: Empirical methods Here's a fun

Linked list loop detection algorithm

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I read some interview question online about how would you find if there's a loop in a linked list, and solution ( Floyd's cycle-finding algorithm ) is to have two pointers, one is 2x faster than the other, and check if they meet again. My question is: Why can't I just keep one pointer fixed, just move the other pointer forward by 1 step each time? Because the first (non-moving) pointer might not lie within the loop, so the pointers would never meet. (Remember that a loop may consist of only part of the list.) Because maybe not the complete linkedList is within the loop. For a linked list lasso

Why increase pointer by two while finding loop in linked list, why not 3,4,5?

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I had a look at question already which talk about algorithm to find loop in a linked list. I have read Floyd's cycle-finding algorithm solution, mentioned at lot of places that we have to take two pointers. One pointer( slower/tortoise ) is increased by one and other pointer( faster/hare ) is increased by 2. When they are equal we find the loop and if faster pointer reaches null there is no loop in the linked list. Now my question is why we increase faster pointer by 2. Why not something else? Increasing by 2 is necessary or we can increase it by X to get the result. Is it necessary that we

Cycle detection in linked list with the Hare and Tortoise approach

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问题 I understand that in order to detect a cycle in a linked list I can use the Hare and Tortoise approach, which holds 2 pointers (slow and fast ones). However, after reading in wiki and other resources, I do not understand why is it guaranteed that the two pointers will meet in O(n) time complexity. 回答1: Here's an attempt at an informal proof. The shape of the cycle doesn't matter. It can have a long tail, and a loop towards the end, or just a loop from the beginning to the end without a tail.

Why increase pointer by two while finding loop in linked list, why not 3,4,5?

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问题 I had a look at question already which talk about algorithm to find loop in a linked list. I have read Floyd's cycle-finding algorithm solution, mentioned at lot of places that we have to take two pointers. One pointer( slower/tortoise ) is increased by one and other pointer( faster/hare ) is increased by 2. When they are equal we find the loop and if faster pointer reaches null there is no loop in the linked list. Now my question is why we increase faster pointer by 2. Why not something else

Explain how finding cycle start node in cycle linked list work?

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I understand that Tortoise and Hare's meeting concludes the existence of loop, but how does moving tortoise to beginning of linked list while keeping the hare at meeting place, followed by moving both one step at a time make them meet at starting point of cycle? This is Floyd's algorithm for cycle detection . You are asking about the second phase of the algorithm -- once you've found a node that's part of a cycle, how does one find the start of the cycle? In the first part of Floyd's algorithm, the hare moves two steps for every step of the tortoise. If the tortoise and hare ever meet, there