[Swift]LeetCode457. 环形数组循环 | Circular Array Loop

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/10343272.html 
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

热烈欢迎，请直接点击！！！

进入博主App Store主页，下载使用各个作品！！！

注：博主将坚持每月上线一个新app！！！

You are given an array of positive and negative integers. If a number n at an index is positive, then move forward n steps. Conversely, if it's negative (-n), move backward n steps. Assume the first element of the array is forward next to the last element, and the last element is backward next to the first element. Determine if there is a loop in this array. A loop starts and ends at a particular index with more than 1 element along the loop. The loop must be "forward" or "backward'.

Example 1: Given the array [2, -1, 1, 2, 2], there is a loop, from index 0 -> 2 -> 3 -> 0.

Example 2: Given the array [-1, 2], there is no loop.

Note: The given array is guaranteed to contain no element "0".

Can you do it in O(n) time complexity and O(1) space complexity?

---

给定一组含有正整数和负整数的数组。如果某个索引中的 n 是正数的，则向前移动 n 个索引。相反，如果是负数(-n)，则向后移动 n 个索引。

假设数组首尾相接。判断数组中是否有环。环中至少包含 2 个元素。环中的元素一律“向前”或者一律“向后”。

示例 1：给定数组 [2, -1, 1, 2, 2], 有一个循环，从索引 0 -> 2 -> 3 -> 0。

示例 2：给定数组[-1, 2], 没有循环。

注意：给定数组保证不包含元素"0"。

你能写出时间复杂度为 O(n) 且空间复杂度为 O(1) 的算法吗？

---

8ms

class Solution {
    func circularArrayLoop(_ nums: [Int]) -> Bool {
        var m:[Int:Int] = [Int:Int]()
        var n:Int = nums.count
        var visited:[Bool] = [Bool](repeating:false,count:n)
        for i in 0..<n
        {
            if visited[i] {continue}
            var cur:Int = i
            while(true)
            {
                visited[cur] = true
                var next:Int = (cur + nums[cur]) % n
                if next < 0 {next += n}
                if next == cur || nums[next] * nums[cur] < 0
                {
                    break
                }
                if m[next] != nil {return true}
                m[cur] = next
                cur = next
            }
        }
        return false
    }
}

---

8ms

class Solution {
    func circularArrayLoop(_ nums: [Int]) -> Bool {
        
        var visitedIndexes: Set<Int> = []
        
        for i in 0..<nums.count {
            if visitedIndexes.contains(i) {
                continue
            }
            
            var currentIndex = i
            var isForwardLoop = nums[currentIndex] > 0
            var lastIndex = -1
            var currentIndexes: Set<Int> = []
            
            while visitedIndexes.count < nums.count {
                visitedIndexes.insert(currentIndex)
                currentIndexes.insert(currentIndex)
                
                let nextIndex = findNextIndex(nums: nums, index: currentIndex)
                
                if isForwardLoop && nums[nextIndex] < 0 {
                    break // we are doing foward loop, and this is backward
                } else if !isForwardLoop && nums[nextIndex] > 0 {
                    break // we are doing backward loop, and this is forward
                } else if nextIndex == lastIndex {
                    // we are looping without 1 number in-between which is fine
                    break
                } else if currentIndexes.contains(nextIndex) {
                    return true // there is a loop
                } else {
                    lastIndex = currentIndex
                    currentIndex = nextIndex
                }
            }
        }
        
        return false
    }
    
    func findNextIndex(nums: [Int], index: Int) -> Int {
        var number = nums[index] % nums.count
        
        var nextIndex = number
        
        if number > 0 {
            nextIndex = (index + number) % nums.count
        } else if number < 0 {
            nextIndex = (index + number)
            if nextIndex < 0 {
                nextIndex += nums.count // wrap it around
            }
        }
        
        return nextIndex
    }
}