[Swift]LeetCode992. K 个不同整数的子数组 | Subarrays with K Different Integers

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Given an array A of positive integers, call a (contiguous, not necessarily distinct) subarray of A good if the number of different integers in that subarray is exactly K.

(For example, [1,2,3,1,2] has 3different integers: 1, 2, and 3.)

Return the number of good subarrays of A.

Example 1:

Input: A = [1,2,1,2,3], K = 2
Output: 7
Explanation: Subarrays formed with exactly 2 different integers: [1,2], [2,1], [1,2], [2,3], [1,2,1], [2,1,2], [1,2,1,2].

Example 2:

Input: A = [1,2,1,3,4], K = 3
Output: 3
Explanation: Subarrays formed with exactly 3 different integers: [1,2,1,3], [2,1,3], [1,3,4].

Note:

1. 1 <= A.length <= 20000
2. 1 <= A[i] <= A.length
3. 1 <= K <= A.length

---

给定一个正整数数组 A，如果 A 的某个子数组中不同整数的个数恰好为 K，则称 A 的这个连续、不一定独立的子数组为好子数组。

（例如，[1,2,3,1,2] 中有 3 个不同的整数：1，2，以及 3。）

返回 A 中好子数组的数目。

示例 1：

输出：A = [1,2,1,2,3], K = 2
输入：7
解释：恰好由 2 个不同整数组成的子数组：[1,2], [2,1], [1,2], [2,3], [1,2,1], [2,1,2], [1,2,1,2].

示例 2：

输入：A = [1,2,1,3,4], K = 3
输出：3
解释：恰好由 3 个不同整数组成的子数组：[1,2,1,3], [2,1,3], [1,3,4].

提示：

1. 1 <= A.length <= 20000
2. 1 <= A[i] <= A.length
3. 1 <= K <= A.length

---

Runtime: 432 ms

Memory Usage: 11.3 MB

class Solution {
    func subarraysWithKDistinct(_ A: [Int], _ K: Int) -> Int {
        return count(A, K) - count(A, K-1)
    }
    
    func count(_ a:[Int],_ K:Int) ->Int
    {
        var n:Int = a.count
        var f:[Int] = [Int](repeating:0,count:n + 1)
        var dis:Int = 0
        var p:Int = 0
        var ret:Int = 0
        for i in 0..<n
        {
            f[a[i]] += 1
            if f[a[i]] == 1
            {
                dis += 1
            }
            while(dis > K)
            {
                f[a[p]] -= 1
                if f[a[p]] == 0
                {
                    dis -= 1
                }
                p += 1
            }
            ret += i-p+1
        }
        return ret
    }
}

---

704ms

class Solution {
    func subarraysWithKDistinct(_ A: [Int], _ K: Int) -> Int {
        var count = 0
        
        var dict1: [Int: Int] = [:]
        var dict2: [Int: Int] = [:]

        var l1 = 0
        var l2 = 0
        
        var r = 0
        
        while r < A.count {
            let num = A[r]
            
            dict1[num, default: 0] += 1
            dict2[num, default: 0] += 1

            while dict1.keys.count > K {
                let l1Num = A[l1]
                dict1[l1Num]! -= 1
                
                if dict1[l1Num] == 0 {
                    dict1[l1Num] = nil
                }
                
                l1 += 1
            }
            
            while dict2.keys.count >= K {
                let l2Num = A[l2]
                dict2[l2Num]! -= 1
                
                if dict2[l2Num] == 0 {
                    dict2[l2Num] = nil
                }
                
                l2 += 1
            }
            
            count += l2 - l1

            r += 1
        }
        
        return count
    }
}