mirror of
https://github.com/halfrost/LeetCode-Go.git
synced 2025-07-05 00:25:22 +08:00
60 lines
3.3 KiB
Markdown
Executable File
60 lines
3.3 KiB
Markdown
Executable File
# [457. Circular Array Loop](https://leetcode.com/problems/circular-array-loop/)
|
||
|
||
|
||
## 题目
|
||
|
||
You are given a **circular** array `nums` of positive and negative integers. If a number k at an index is positive, then move forward k steps. Conversely, if it's negative (-k), move backward k steps. Since the array is circular, you may assume that the last element's next element is the first element, and the first element's previous element is the last element.
|
||
|
||
Determine if there is a loop (or a cycle) in `nums`. A cycle must start and end at the same index and the cycle's length > 1. Furthermore, movements in a cycle must all follow a single direction. In other words, a cycle must not consist of both forward and backward movements.
|
||
|
||
**Example 1:**
|
||
|
||
Input: [2,-1,1,2,2]
|
||
Output: true
|
||
Explanation: There is a cycle, from index 0 -> 2 -> 3 -> 0. The cycle's length is 3.
|
||
|
||
**Example 2:**
|
||
|
||
Input: [-1,2]
|
||
Output: false
|
||
Explanation: The movement from index 1 -> 1 -> 1 ... is not a cycle, because the cycle's length is 1. By definition the cycle's length must be greater than 1.
|
||
|
||
**Example 3:**
|
||
|
||
Input: [-2,1,-1,-2,-2]
|
||
Output: false
|
||
Explanation: The movement from index 1 -> 2 -> 1 -> ... is not a cycle, because movement from index 1 -> 2 is a forward movement, but movement from index 2 -> 1 is a backward movement. All movements in a cycle must follow a single direction.
|
||
|
||
**Note:**
|
||
|
||
1. -1000 ≤ nums[i] ≤ 1000
|
||
2. nums[i] ≠ 0
|
||
3. 1 ≤ nums.length ≤ 5000
|
||
|
||
**Follow up:**
|
||
|
||
Could you solve it in **O(n)** time complexity and **O(1)** extra space complexity?
|
||
|
||
|
||
## 题目大意
|
||
|
||
给定一个含有正整数和负整数的环形数组 nums。 如果某个索引中的数 k 为正数，则向前移动 k 个索引。相反，如果是负数 (-k)，则向后移动 k 个索引。因为数组是环形的，所以可以假设最后一个元素的下一个元素是第一个元素，而第一个元素的前一个元素是最后一个元素。
|
||
|
||
确定 nums 中是否存在循环（或周期）。循环必须在相同的索引处开始和结束并且循环长度 > 1。此外，一个循环中的所有运动都必须沿着同一方向进行。换句话说，一个循环中不能同时包括向前的运动和向后的运动。
|
||
|
||
提示：
|
||
|
||
- -1000 ≤ nums[i] ≤ 1000
|
||
- nums[i] ≠ 0
|
||
- 1 ≤ nums.length ≤ 5000
|
||
|
||
进阶：
|
||
|
||
- 你能写出时间时间复杂度为 O(n) 和额外空间复杂度为 O(1) 的算法吗？
|
||
|
||
## 解题思路
|
||
|
||
|
||
- 给出一个循环数组，数组的数字代表了前进和后退的步数，+ 代表往右(前进)，- 代表往左(后退)。问这个循环数组中是否存在一个循环，并且这个循环内不能只有一个元素，循环的方向都必须是同方向的。
|
||
- 遇到循环就可以优先考虑用快慢指针的方法判断循环，这一题对循环增加了一个条件，循环不能只是单元素的循环，所以在快慢指针中加入这个判断条件。还有一个判断条件是循环的方向必须是同向的，这个简单，用 `num[i] * num[j] > 0` 就可以判断出是同向的(如果是反向的，那么两者的乘积必然是负数)，如果没有找到循环，可以将当前已经走过的路径上的 num[] 值都置为 0，标记已经访问过了。下次循环遇到访问过的元素，`num[i] * num[j] > 0` 就会是 0，提前退出找循环的过程。
|