[Swift]LeetCode930. 和相同的二元子数组 | Binary Subarrays With Sum

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In an array A of 0s and 1s, how many non-empty subarrays have sum S?

 Example 1:

Input: A = [1,0,1,0,1], S = 2
Output: 4
Explanation: 
The 4 subarrays are bolded below:
[1,0,1,0,1]
[1,0,1,0,1]
[1,0,1,0,1]
[1,0,1,0,1]

 Note:

1. A.length <= 30000
2. 0 <= S <= A.length
3. A[i] is either 0 or 1.

---

在由若干 0 和 1  组成的数组 A 中，有多少个和为 S 的非空子数组。

 示例：

输入：A = [1,0,1,0,1], S = 2
输出：4
解释：
如下面黑体所示，有 4 个满足题目要求的子数组：
[1,0,1,0,1]
[1,0,1,0,1]
[1,0,1,0,1]
[1,0,1,0,1]

 提示：

1. A.length <= 30000
2. 0 <= S <= A.length
3. A[i] 为 0 或 1

---

220ms

class Solution {
    func numSubarraysWithSum(_ A: [Int], _ S: Int) -> Int {
        var n:Int = A.count
        var cum:[Int] = [Int](repeating: 0,count: n + 1)
        for i in 0..<n
        {
            cum[i + 1] = cum[i] + A[i]
        }
        
        var ret:Int = 0
        var f:[Int] = [Int](repeating: 0,count: 30003)
        for i in 0...n
        {
            if cum[i] - S >= 0
            {
                ret += f[cum[i] - S]
            }
            f[cum[i]] += 1
        }
        return ret
    }
}

---

232ms

class Solution {
    func numSubarraysWithSum(_ A: [Int], _ S: Int) -> Int {
        
        var count = 0
        var zCount = 0
        var prevVal = 0
        if S == 0 {
            for val in A {
                if val == 0 {
                    print(zCount)
                    prevVal += 1
                    zCount += prevVal
                } else {
                    count += zCount
                    zCount = 0
                    prevVal = 0
                }
            }
            count += zCount
            return count
        }
        
        var leftZeros:[Int] = Array(repeating: 0, count: A.count)
        var rightZeros:[Int] = Array(repeating: 0, count: A.count)
        var onesInd:[Int] = []
        
        
        for i in 0..<A.count {
            if A[i] == 0 {
                count += 1
            } else {
            leftZeros[i] = count
            onesInd.append(i)
            count = 0
            }
        }
        
        count = 0
        for i in (0..<A.count).reversed() {
            if A[i] == 0 {
                count += 1
            }else {
            rightZeros[i] = count
            count = 0
            }
        }
        
        if onesInd.count < S {
            return 0
        }
        
        count = 0
        for i in 0...(onesInd.count - S) {
            let leftInd = onesInd[i]
            let rightInd = onesInd[i + S - 1]
            let subCount = (1 + leftZeros[leftInd]) * (rightZeros[rightInd] + 1)
            count += subCount
        }
        return count
    }
}

---

304ms

class Solution {
    func numSubarraysWithSum(_ A: [Int], _ S: Int) -> Int {
        var counter = 0
        var dp = [Int](repeating: 0, count: A.count + 1)
        for i in 0..<A.count {
            counter += A[i]
            dp[i + 1] = counter
        }
        var result = 0
        var dict = [Int : Int]()
        dict[0] = 1
        for i in 1..<dp.count {
            let a = dp[i]
            if let value = dict[a - S] {
                result += value
            }
            dict[a] = dict[a] ?? 0
            dict[a] = dict[a]! + 1
        }
        return result 
    }
}