[Swift]LeetCode565. 数组嵌套 | Array Nesting

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A zero-indexed array A of length N contains all integers from 0 to N-1. Find and return the longest length of set S, where S[i] = {A[i], A[A[i]], A[A[A[i]]], ... } subjected to the rule below.

Suppose the first element in S starts with the selection of element A[i] of index = i, the next element in S should be A[A[i]], and then A[A[A[i]]]… By that analogy, we stop adding right before a duplicate element occurs in S. 

Example 1:

Input: A = [5,4,0,3,1,6,2]
Output: 4
Explanation: 
A[0] = 5, A[1] = 4, A[2] = 0, A[3] = 3, A[4] = 1, A[5] = 6, A[6] = 2.

One of the longest S[K]:
S[0] = {A[0], A[5], A[6], A[2]} = {5, 6, 2, 0} 

Note:

1. N is an integer within the range [1, 20,000].
2. The elements of A are all distinct.
3. Each element of A is an integer within the range [0, N-1].

---

索引从0开始长度为N的数组A，包含0到N - 1的所有整数。找到并返回最大的集合S，S[i] = {A[i], A[A[i]], A[A[A[i]]], ... }且遵守以下的规则。

假设选择索引为i的元素A[i]为S的第一个元素，S的下一个元素应该是A[A[i]]，之后是A[A[A[i]]]... 以此类推，不断添加直到S出现重复的元素。

示例 1:

输入: A = [5,4,0,3,1,6,2]
输出: 4
解释: 
A[0] = 5, A[1] = 4, A[2] = 0, A[3] = 3, A[4] = 1, A[5] = 6, A[6] = 2.

其中一种最长的 S[K]:
S[0] = {A[0], A[5], A[6], A[2]} = {5, 6, 2, 0}

注意:

1. N是[1, 20,000]之间的整数。
2. A中不含有重复的元素。
3. A中的元素大小在[0, N-1]之间。

---

Runtime: 124 ms

Memory Usage: 19.5 MB

class Solution {
    func arrayNesting(_ nums: [Int]) -> Int {
        var nums = nums
        var n:Int = nums.count
        var res:Int = 0
        for i in 0..<n
        {
            var cnt:Int = 1
            while(nums[i] != i && nums[i] != nums[nums[i]])
            {
                nums.swapAt(i,nums[i])
                cnt += 1
            }
            res = max(res,cnt)
        }
        return res        
    }
}

---

140ms

class Solution {
    func arrayNesting(_ nums: [Int]) -> Int {
        var visited: Set<Int> = []
        var maxLen = 0
        for i in 0 ..< nums.count {
            if !visited.contains(i) {
                var next = nums[i]
                var count = 0
                while !visited.contains(next) {
                    visited.insert(next)
                    next = nums[next]
                    count += 1
                }
                maxLen = max(maxLen, count)
            }
        }
        
        return maxLen
    }
}

---

184ms

class Solution {
    func arrayNesting(_ nums: [Int]) -> Int {
        var visited = [Bool](repeating: false, count: nums.count)
        var result = 0
        
        for i in 0..<nums.count {
            var j = i
            var count = 0
            while !visited[j] {
                visited[j] = true
                j = nums[j]
                count += 1
            }
            result = max(result, count)
        }
        
        return result
    }
}