• [Swift]LeetCode795. 区间子数组个数 | Number of Subarrays with Bounded Maximum


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    We are given an array A of positive integers, and two positive integers L and R (L <= R).

    Return the number of (contiguous, non-empty) subarrays such that the value of the maximum array element in that subarray is at least L and at most R.

    Example :
    Input: 
    A = [2, 1, 4, 3]
    L = 2
    R = 3
    Output: 3
    Explanation: There are three subarrays that meet the requirements: [2], [2, 1], [3].
    

    Note:

    • L, R  and A[i] will be an integer in the range [0, 10^9].
    • The length of A will be in the range of [1, 50000].

    给定一个元素都是正整数的数组A ,正整数 L 以及 R (L <= R)。

    求连续、非空且其中最大元素满足大于等于L 小于等于R的子数组个数。

    例如 :
    输入: 
    A = [2, 1, 4, 3]
    L = 2
    R = 3
    输出: 3
    解释: 满足条件的子数组: [2], [2, 1], [3].
    

    注意:

    • L, R  和 A[i] 都是整数,范围在 [0, 10^9]
    • 数组 A 的长度范围在[1, 50000]

    Runtime: 256ms

    Memory Usage: 19.6 MB
     1 class Solution {
     2     func numSubarrayBoundedMax(_ A: [Int], _ L: Int, _ R: Int) -> Int {
     3         var res:Int = 0
     4         var left:Int = -1
     5         var right:Int = -1
     6         for i in 0..<A.count
     7         {
     8             if A[i] > R
     9             {
    10                 left = i
    11                 right = i
    12                 continue
    13             }
    14             if A[i] >= L
    15             {
    16                 right = i
    17             }
    18             res += right - left
    19         }
    20         return res
    21     }
    22 }

    Runtime: 288 ms
    Memory Usage: 19.6 MB
     1 class Solution {
     2     func numSubarrayBoundedMax(_ A: [Int], _ L: Int, _ R: Int) -> Int {
     3          var first = 0,second = 0
     4         var result = 0,rangeCount = 0
     5         
     6         while second < A.count && first < A.count {
     7             if A[second] >= L && A[second] <= R {
     8                 result += (second - first + 1)
     9                 rangeCount = second - first + 1
    10             } else if (A[second] < L) {
    11                 result += rangeCount
    12             } else { // A[second] > R
    13                 first = second + 1
    14                 rangeCount = 0
    15             }
    16             second += 1
    17         }
    18         return result
    19     }    
    20 }

    300ms

     1 class Solution {
     2     func numSubarrayBoundedMax(_ A: [Int], _ L: Int, _ R: Int) -> Int {
     3         var result = 0
     4         var numberOfLows = 0
     5         var numberOfEquals = 0
     6         enum Status {
     7             case equal
     8             case low
     9             case high
    10         }
    11         var status = Status.equal
    12         
    13         for (_, value) in A.enumerated() {
    14             switch status {
    15             case .equal:
    16                 if value < L {
    17                     numberOfLows = 1
    18                     numberOfEquals += 1
    19                 } else if value > R {
    20                     result += numberOfSubarrayInLength(numberOfEquals)
    21                     numberOfEquals = 0
    22                 } else {
    23                     numberOfEquals += 1
    24                 }
    25             case .low:
    26                 if value < L {
    27                     numberOfLows += 1
    28                     numberOfEquals += 1
    29                 } else if value > R {
    30                     result += numberOfSubarrayInLength(numberOfEquals)
    31                     numberOfEquals = 0
    32                     result -= numberOfSubarrayInLength(numberOfLows)
    33                     numberOfLows = 0
    34                 } else {
    35                     numberOfEquals += 1
    36                     result -= numberOfSubarrayInLength(numberOfLows)
    37                     numberOfLows = 0
    38                 }
    39             case .high:
    40                 if value < L {
    41                     numberOfLows = 1
    42                     numberOfEquals = 1
    43                 } else if value > R {
    44                     
    45                 } else {
    46                     numberOfLows = 0
    47                     numberOfEquals = 1
    48                 }
    49             }
    50             if value < L {
    51                 status = .low
    52             } else if value > R {
    53                 status = .high
    54             } else {
    55                 status = .equal
    56             }
    57         }
    58         result -= numberOfSubarrayInLength(numberOfLows)
    59         result += numberOfSubarrayInLength(numberOfEquals)
    60         return result
    61     }
    62     
    63     func numberOfSubarrayInLength(_ n: Int) -> Int {
    64         if n <= 0 {
    65             return 0
    66         }
    67         
    68         return n * (n + 1) / 2
    69     }
    70 }

    356ms

     1 class Solution {
     2     func numSubarrayBoundedMax(_ A: [Int], _ L: Int, _ R: Int) -> Int {
     3         var count = 0
     4         for i in 0..<A.count {
     5             var currMax = A[i]
     6             for j in i..<A.count {
     7                 currMax = max(currMax, A[j])
     8                 if currMax >= L && currMax <= R {
     9                     count += 1
    10                 } else if currMax > R {
    11                     break
    12                 }
    13             }
    14         }
    15         return count
    16     }
    17 }

    3816ms

     1 class Solution {
     2     func numSubarrayBoundedMax(_ A: [Int], _ L: Int, _ R: Int) -> Int {
     3         if A.count <= 0 { return 0 }
     4         var cursor = 0
     5         var count = 0
     6         while cursor < A.count {
     7             for i in cursor ..< A.count {
     8                 let temp = A[cursor...i]
     9                 let max = temp.max()!
    10                 if max > R {
    11                     break
    12                 }
    13                 if max >= L {
    14                     count += 1
    15                 }
    16             }
    17             cursor += 1
    18         }
    19         return count
    20     }
    21 }
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  • 原文地址:https://www.cnblogs.com/strengthen/p/10547191.html
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