• Home
• com.github.javadev
• leetcode-in-java
• 1.23
• source code
• readme.md

×

Project download

Many resources are needed to download a project. Please understand that we have to compensate our server costs. Thank you in advance.
Project price only 1 $

You can buy this project and download/modify it how often you want.

g0401_0500.s0457_circular_array_loop.readme.md Maven / Gradle / Ivy

457\. Circular Array Loop

Medium

You are playing a game involving a **circular** array of non-zero integers `nums`. Each `nums[i]` denotes the number of indices forward/backward you must move if you are located at index `i`:

*   If `nums[i]` is positive, move `nums[i]` steps **forward**, and
*   If `nums[i]` is negative, move `nums[i]` steps **backward**.

Since the array is **circular**, you may assume that moving forward from the last element puts you on the first element, and moving backwards from the first element puts you on the last element.

A **cycle** in the array consists of a sequence of indices `seq` of length `k` where:

*   Following the movement rules above results in the repeating index sequence `seq[0] -> seq[1] -> ... -> seq[k - 1] -> seq[0] -> ...`
*   Every `nums[seq[j]]` is either **all positive** or **all negative**.
*   `k > 1`

Return `true` _if there is a **cycle** in_ `nums`_, or_ `false` _otherwise_.

**Example 1:**

**Input:** nums = [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:** nums = [-1,2]

**Output:** false

**Explanation:** The sequence from index 1 -> 1 -> 1 -> ... is not a cycle because the sequence's length is 1. By definition the sequence's length must be strictly greater than 1 to be a cycle.

**Example 3:**

**Input:** nums = [-2,1,-1,-2,-2]

**Output:** false

**Explanation:** The sequence from index 1 -> 2 -> 1 -> ... is not a cycle because nums[1] is positive, but nums[2] is negative. Every nums[seq[j]] must be either all positive or all negative.

**Constraints:**

*   `1 <= nums.length <= 5000`
*   `-1000 <= nums[i] <= 1000`
*   `nums[i] != 0`

**Follow up:** Could you solve it in `O(n)` time complexity and `O(1)` extra space complexity?