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

发布时间：2020-12-14 05:03:58 所属栏目：百科来源：网络整理

导读：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 = 2Output: 2? Note: The length of the array is in range [1,20,000]. The range of

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:

The length of the array is in range [1,20,000].
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,k = 2
输出: 2,[1,1] 与 [1,1] 为两种不同的情况。

说明 :

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

56ms

 1 class Solution {
 2     func subarraySum(_ nums: [Int],_ k: Int) -> Int {
 3                 if nums.count == 0 {
 4             return 0
 5         }
 6         var dic : Dictionary<Int,Int> = [:]
 7         var sum : Int = 0;
 8         var res : Int = 0;
 9         for i in 0 ..< nums.count {
10             sum += nums[i]
11             
12             if sum == k {
13                 res += 1
14             } 
15             
16             if dic[sum - k] != nil {
17                 res += dic[sum - k]!
18             }
19             
20             if dic[sum] == nil {
21                 dic[sum] = 1
22             } else {
23                 dic[sum] = dic[sum]! + 1
24             }
25         }        
26         return res
27     }
28 }

120ms

 1 class Solution {
 2     func subarraySum(_ nums: [Int],_ k: Int) -> Int {
 3         var sumCounter = [0: 1]
 4         var s = 0,res = 0
 5         for i in 0..<nums.count {
 6             s += nums[i]
 7             res += sumCounter[s - k] ?? 0
 8             sumCounter[s,default: 0] += 1
 9         }
10         return res
11     }
12 }

Runtime:?124 ms

Memory Usage:?19.8 MB

 1 class Solution {
 2     func subarraySum(_ nums: [Int],_ k: Int) -> Int {
 3         var res:Int = 0
 4         var sum:Int = 0
 5         var n:Int = nums.count
 6         var m:[Int:Int] = [0:1]
 7         for i in 0..<n
 8         {
 9             sum += nums[i]
10             res += m[sum - k,default:0]
11             m[sum,default:0] += 1
12         }
13         return res
14     }
15 }

124ms

 1 class Solution {
 2     func subarraySum(_ nums: [Int],_ k: Int) -> Int {
 3       
 4       guard nums.count > 0 else { return -1 }
 5         var count = 0
 6         var sum = 0
 7         
 8         var dictionary: [Int: Int] = [:]
 9         dictionary[0] = 1
10         for i in 0..<nums.count {
11             sum += nums[i]
12             if let occurance = dictionary[sum - k] {
13                 count += occurance
14             }
15             
16             if let occurance = dictionary[sum] {
17                 dictionary[sum] = occurance + 1
18             } else {
19                 dictionary[sum] = 1
20             }
21             
22         }
23         
24         return count
25     }
26 }

148ms

 1 class Solution {
 2     func subarraySum(_ nums: [Int],_ k: Int) -> Int {
 3         var partialSums = [Int: [Int]]()
 4         partialSums[0] = [0]
 5         var found = [(Int,Int)]()
 6         var sum = 0
 7         for (i,num) in nums.enumerated() {
 8             sum += num
 9             if let bounds = partialSums[sum - k] {
10                 bounds.forEach { found.append(($0,i)) }
11             }
12             partialSums[sum,default: []].append(i)
13         }
14         return found.count
15     }
16 }

（编辑：李大同）

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