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[Leetcode with JavaScript] 刷題筆記 — 特殊方法 Floyd's Cycle Detection

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Justin Huang
作者
Justin Huang
Senior Software Engineer@Delta Energy

前言
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這是最近刷 Leetcode 學習到的特殊方法,適合拿來解尋找重複數字的題目, 如 Array 題型 287. Find the Duplicate Number,還能解答 Linked List 相關題型如 141. Linked List Cycle、142. Linked List Cycle II,覺得相當實用,因此決定寫成筆記,內容著重於應用在刷題,若想細究原理的讀者可以參考底下的 Reference #4 #5。

Floyd’s Cycle Detection
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有發現封面圖片上出現了烏龜與兔子嗎?因為這個演算法俗稱龜兔賽跑演算法。以下是原理及結論:

  1. 假設數列是一個賽道,龜兔一起出發,兔子跑得比烏龜快,如果這個賽道中沒有環(cycle)存在,在龜兔速度不變的情況下,烏龜永遠不會與兔子相遇;但***只要賽道中有環,那龜兔一定會在某一點相遇*** (可參考 Reference #1)
  2. 龜兔相遇後,可藉由將烏龜退回起點,兔子在原地不動,此時龜兔之間的距離,會是環的整數倍,然後讓龜兔每次只動一格,兩者交會時就會是環的起點。

龜兔兩個在數列裡跑,聽起來好像有點熟悉,沒錯,這個概念和我在 上一篇刷題筆記中所介紹的雙指針 Two Pointers 方法很像,只是指針一個快一個慢,因此又稱為「快慢指針」。

接下來直接看題目實作吧!

287. Find the Duplicate Number
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Given an array of integers nums containing n + 1 integers where each integer is in the range [1, n] inclusive.

There is only one duplicate number in nums, return this duplicate number.

Follow-ups:

  1. How can we prove that at least one duplicate number must exist in nums?
  2. Can you solve the problem without modifying the array nums?
  3. Can you solve the problem using only constant, O(1) extra space?
  4. Can you solve the problem with runtime complexity less than O(n2)?

Example 1:

Input: nums = [1,3,4,2,2]
Output: 2

Example 2:

Input: nums = [3,1,3,4,2]
Output: 3

題目要求是找出陣列中重複的數字然後回傳,聽起來好像不難嘛 😏,但是最後還有但書,空間複雜度 O(1),時間複雜度 O(n²) 😱,因此若想用暴力法搭配 .sort() 都不符合題目的規範。

這時候就需要使用快慢指針了,以下是思考方式:

  1. 使用兩個變數 slow fast 來代表慢指針(烏龜) 及 快指針(兔子),他們一開始會在原點也就是 nums[0]
  2. 然後讓龜兔速度不同,因此 fast會用兩層陣列的方式,來實作兔子走得比較快的狀況
  3. 利用迴圈讓龜兔重複 #2 的動作,直到 **slow** **fast**相等(龜兔相遇),代表有環存在
  4. slow退回原點nums[0]fast保持在原地不動
  5. 然後再次利用迴圈讓龜兔開始跑,但這次龜兔的速度要是一致的,再次相遇時,就是環的起點,也代表重複的數字
var findDuplicate = function (nums) {
  let slow = nums[0];
  let fast = nums[0];

  slow = nums[slow];
  fast = nums[nums[fast]];

  // find meeting point
  while (slow !== fast) {
    slow = nums[slow];
    fast = nums[nums[fast]];
  }

  // find entry point of cycle
  slow = nums[0];
  while (slow !== fast) {
    slow = nums[slow]; // slow back to start point
    fast = nums[fast]; // fast stay at meeting point
  }
  return fast;
};

這個方法也能應用在 Linked List 的題型,先來題簡單的:

141. Linked List Cycle
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Given a linked list, determine if it has a cycle in it.

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 the tail connects to. If pos == -1, then there is no cycle in the linked list.

Follow up:

Can you solve it using O(1) (i.e. constant) memory?

Example 1:

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Input: head = [3,2,0,-4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the second node.

這題對空間複雜度有相同的要求,因此也相當適合使用快慢指針,基本上邏輯相同,就不再贅述了,如果有學過 Linked List 的人可以順便練習一下。

var hasCycle = function (head) {
  let fast = head;
  let slow = head;

  // check the cycle
  while (fast && fast.next) {
    fast = fast.next.next;
    slow = slow.next;
    // cycle exist
    if (fast === slow) return true;
  }
  // no cycle
  return false;
};

142. Linked List Cycle II
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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.

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var detectCycle = function (head) {
  let fast = head;
  let slow = head;
  while (fast && fast.next) {
    fast = fast.next.next;
    slow = slow.next;
    if (fast === slow) {
      slow = head;
      while (fast !== slow) {
        fast = fast.next;
        slow = slow.next;
      }
      return slow;
    }
  }
  return null;
};

Reference
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  1. If Programming Was An Anime
  2. 判圈算法
  3. 找出重複數字:Floyd Cycle Detection Algorithm 龜兔賽跑演算法
  4. 探索 Floyd Cycle Detection Algorithm
  5. [ Wiki] Cycle detection