[Swift]LeetCode713. 乘积小于K的子数组 | Subarray Product Less Than K

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤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/10506745.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！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

Your are given an array of positive integers nums.

Count and print the number of (contiguous) subarrays where the product of all the elements in the subarray is less than k.

Example 1:

Input: nums = [10, 5, 2, 6], k = 100
Output: 8
Explanation: The 8 subarrays that have product less than 100 are: [10], [5], [2], [6], [10, 5], [5, 2], [2, 6], [5, 2, 6].
Note that [10, 5, 2] is not included as the product of 100 is not strictly less than k.

Note:

• 0 < nums.length <= 50000.
• 0 < nums[i] < 1000.
• 0 <= k < 10^6.

---

给定一个正整数数组 nums。

找出该数组内乘积小于 k 的连续的子数组的个数。

示例 1:

输入: nums = [10,5,2,6], k = 100
输出: 8
解释: 8个乘积小于100的子数组分别为: [10], [5], [2], [6], [10,5], [5,2], [2,6], [5,2,6]。
需要注意的是 [10,5,2] 并不是乘积小于100的子数组。

说明:

• 0 < nums.length <= 50000
• 0 < nums[i] < 1000
• 0 <= k < 10^6

---

 872ms

class Solution {
    func numSubarrayProductLessThanK(_ nums: [Int], _ k: Int) -> Int {
        guard nums.count > 0 else {
            return 0 
        }
        
        var movingIndex: Int = 0 
        var startIndex: Int = 0 
        var product: Int = 1 
        var result: Int = 0 
        
        while movingIndex < nums.count {
            let num = nums[movingIndex]
            product *= num 
             while product >= k && startIndex < movingIndex {
                    product /= nums[startIndex]
                    startIndex += 1
            }
            if product < k {
                result += movingIndex - startIndex + 1
            }
             movingIndex += 1           
        }
        
        return result
    }
}

---

Runtime: 880 ms

Memory Usage: 19 MB

class Solution {
    func numSubarrayProductLessThanK(_ nums: [Int], _ k: Int) -> Int {
        if k <= 1 {return 0}
        var res:Int = 0
        var prod:Int = 1
        var left:Int = 0
        for i in 0..<nums.count
        {
            prod *= nums[i]
            while(prod >= k)
            {
                prod /= nums[left]
                left += 1
            }
            res += i - left + 1
        }
        return res
    }
}

---

896ms

class Solution {
    func numSubarrayProductLessThanK(_ nums: [Int], _ k: Int) -> Int {

    var left: Int = 0
    var sum = 1,totle = 0,right = 0;
    
    if k <= sum {
        return 0
    }
    
    for (index, value) in nums.enumerated() {
        
        sum *= value;
        while sum >= k {
            sum = sum/nums[left]
            left += 1
        }
        totle += ((right - left) + 1)
        right += 1
        
    }
    return totle;
  }
}