[Swift]LeetCode560. 和为K的子数组 | Subarray Sum Equals K

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Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k.

Example 1:

Input:nums = [1,1,1], k = 2
Output: 2 

Note:

1. The length of the array is in range [1, 20,000].
2. The range of numbers in the array is [-1000, 1000] and the range of the integer k is [-1e7, 1e7].

---

给定一个整数数组和一个整数 k，你需要找到该数组中和为 k 的连续的子数组的个数。

示例 1 :

输入:nums = [1,1,1], k = 2
输出: 2 , [1,1] 与 [1,1] 为两种不同的情况。

说明 :

1. 数组的长度为 [1, 20,000]。
2. 数组中元素的范围是 [-1000, 1000] ，且整数 k 的范围是 [-1e7, 1e7]。

---

56ms

class Solution {
    func subarraySum(_ nums: [Int], _ k: Int) -> Int {
                if nums.count == 0 {
            return 0
        }
        var dic : Dictionary<Int, Int> = [:]
        var sum : Int = 0;
        var res : Int = 0;
        for i in 0 ..< nums.count {
            sum += nums[i]
            
            if sum == k {
                res += 1
            } 
            
            if dic[sum - k] != nil {
                res += dic[sum - k]!
            }
            
            if dic[sum] == nil {
                dic[sum] = 1
            } else {
                dic[sum] = dic[sum]! + 1
            }
        }        
        return res
    }
}

---

120ms

class Solution {
    func subarraySum(_ nums: [Int], _ k: Int) -> Int {
        var sumCounter = [0: 1]
        var s = 0, res = 0
        for i in 0..<nums.count {
            s += nums[i]
            res += sumCounter[s - k] ?? 0
            sumCounter[s, default: 0] += 1
        }
        return res
    }
}

---

Runtime: 124 ms

Memory Usage: 19.8 MB

class Solution {
    func subarraySum(_ nums: [Int], _ k: Int) -> Int {
        var res:Int = 0
        var sum:Int = 0
        var n:Int = nums.count
        var m:[Int:Int] = [0:1]
        for i in 0..<n
        {
            sum += nums[i]
            res += m[sum - k,default:0]
            m[sum,default:0] += 1
        }
        return res
    }
}

---

124ms

class Solution {
    func subarraySum(_ nums: [Int], _ k: Int) -> Int {
      
      guard nums.count > 0 else { return -1 }
        var count = 0
        var sum = 0
        
        var dictionary: [Int: Int] = [:]
        dictionary[0] = 1
        for i in 0..<nums.count {
            sum += nums[i]
            if let occurance = dictionary[sum - k] {
                count += occurance
            }
            
            if let occurance = dictionary[sum] {
                dictionary[sum] = occurance + 1
            } else {
                dictionary[sum] = 1
            }
            
        }
        
        return count
    }
}

---

148ms

class Solution {
    func subarraySum(_ nums: [Int], _ k: Int) -> Int {
        var partialSums = [Int: [Int]]()
        partialSums[0] = [0]
        var found = [(Int, Int)]()
        var sum = 0
        for (i, num) in nums.enumerated() {
            sum += num
            if let bounds = partialSums[sum - k] {
                bounds.forEach { found.append(($0, i)) }
            }
            partialSums[sum, default: []].append(i)
        }
        return found.count
    }
}