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  • [Swift]LeetCode523. 连续的子数组和 | Continuous Subarray Sum

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    ➤原文地址:https://www.cnblogs.com/strengthen/p/10397532.html 
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    Given a list of non-negative numbers and a target integer k, write a function to check if the array has a continuous subarray of size at least 2 that sums up to the multiple of k, that is, sums up to n*k where n is also an integer.

    Example 1:

    Input: [23, 2, 4, 6, 7],  k=6
    Output: True
    Explanation: Because [2, 4] is a continuous subarray of size 2 and sums up to 6. 

    Example 2:

    Input: [23, 2, 6, 4, 7],  k=6
    Output: True
    Explanation: Because [23, 2, 6, 4, 7] is an continuous subarray of size 5 and sums up to 42. 

    Note:

    1. The length of the array won't exceed 10,000.
    2. You may assume the sum of all the numbers is in the range of a signed 32-bit integer.

    给定一个包含非负数的数组和一个目标整数 k,编写一个函数来判断该数组是否含有连续的子数组,其大小至少为 2,总和为 k 的倍数,即总和为 n*k,其中 n 也是一个整数。

    示例 1:

    输入: [23,2,4,6,7], k = 6
    输出: True
    解释: [2,4] 是一个大小为 2 的子数组,并且和为 6。
    

    示例 2:

    输入: [23,2,6,4,7], k = 6
    输出: True
    解释: [23,2,6,4,7]是大小为 5 的子数组,并且和为 42。
    

    说明:

    1. 数组的长度不会超过10,000。
    2. 你可以认为所有数字总和在 32 位有符号整数范围内。

    192ms

     1 class Solution {
     2     func checkSubarraySum(_ nums: [Int], _ k: Int) -> Bool {
     3         var map = [Int: Int]()
     4         map[0] = -1
     5         var runningSum = 0
     6         for i in 0..<nums.count {
     7             runningSum += nums[i]
     8             if k != 0 {
     9                 runningSum %= k
    10             }
    11             let pre = map[runningSum]
    12             if pre != nil {
    13                 if (i - pre!) > 1 { return true}
    14             } else {
    15                 map[runningSum] = i
    16             }
    17         }
    18         return false
    19     }
    20 }

    232ms

     1 class Solution {
     2     func checkSubarraySum(_ nums: [Int], _ k: Int) -> Bool {
     3         if nums.count < 2 { return false }       
     4         if (k == 0) { return nums[0] == 0 && nums[1] == 0 }
     5         var dict = [Int:Int]()
     6         
     7         var sum = 0
     8         for i in 0...nums.count-1 {
     9             sum += nums[i]
    10             
    11             let r = sum % k 
    12             if r == 0 { 
    13                 
    14                 if(i>=1) {
    15                     
    16                     return true
    17                 }
    18             }
    19             if let value = dict[r] {
    20                 if (i - value >= 2) { return true }
    21             }
    22             else {
    23                 dict[r] = i
    24             }
    25             
    26             
    27         }
    28         return false
    29     }
    30 }

    316ms

     1 class Solution {
     2     func isMultiple(_ num: Int, _ k: Int) -> Bool {
     3         if k == 0 {
     4             return num == 0
     5         } else {
     6             return num % k == 0
     7         }
     8     }
     9     func checkSubarraySum(_ nums: [Int], _ k: Int) -> Bool {
    10         var preSum: [Int] = []
    11         for (i, num) in nums.enumerated() {
    12             let sum = (preSum.last ?? 0) + num
    13             preSum.append(sum)
    14             if i >= 1 && isMultiple(sum, k) {
    15                 return true
    16             }
    17         }
    18         if 0 > nums.count - 2 {
    19             return false
    20         }
    21         for i in 0..<(nums.count - 2) {
    22             for j in (i + 2)..<nums.count {
    23                 let sum = preSum[j] - preSum[i]
    24                 if isMultiple(sum, k) {
    25                     return true
    26                 }
    27             }
    28         }
    29         return false
    30     }
    31 }

    324ms

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

    340ms

     1 class Solution {
     2     func checkSubarraySum(_ nums: [Int], _ k: Int) -> Bool {
     3         var sum: [Int] = []
     4         var total = 0
     5         for i in 0..<nums.count {
     6             total += nums[i]
     7             sum.append(total)
     8         }
     9         
    10         for start in 0..<nums.count-1 {
    11             for end in start+1..<nums.count {
    12                 let summ = sum[end] - sum[start] + nums[start]
    13                 if summ == k || (k != 0 && summ%k == 0) {
    14                     return true
    15                 }
    16             }
    17         }
    18         
    19         return false
    20     }
    21 }

    380ms

     1 class Solution {
     2     func checkSubarraySum(_ nums: [Int], _ k: Int) -> Bool {
     3     return checkSubarraySumRecursion(nums, k)
     4 }
     5 
     6 func checkSubarraySumRecursion(_ nums : [Int], _ k : Int)->Bool{
     7     guard nums.count > 1 else{
     8         return false
     9     }
    10     
    11     var sum : Int = nums[0]
    12     for index in 1..<nums.count{
    13         sum += nums[index]
    14         if k == 0 && sum == 0{
    15             return true
    16         }
    17         
    18         if k != 0 && sum % k == 0{
    19             return true
    20         }
    21     }
    22     
    23     return checkSubarraySumRecursion(Array(nums[1...]), k)
    24   }
    25 }
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  • 原文地址:https://www.cnblogs.com/strengthen/p/10397532.html
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