zoukankan html css js c++ java

[Swift]LeetCode532. 数组中的K-diff数对 | K-diff Pairs in an Array

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/9813030.html
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

Given an array of integers and an integer k, you need to find the number of unique k-diff pairs in the array. Here a k-diff pair is defined as an integer pair (i, j), where i and j are both numbers in the array and their absolute difference is k.

Example 1:

Input: [3, 1, 4, 1, 5], k = 2
Output: 2
Explanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5).
Although we have two 1s in the input, we should only return the number of unique pairs.

Example 2:

Input:[1, 2, 3, 4, 5], k = 1
Output: 4
Explanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5).

Example 3:

Input: [1, 3, 1, 5, 4], k = 0
Output: 1
Explanation: There is one 0-diff pair in the array, (1, 1).

Note:

The pairs (i, j) and (j, i) count as the same pair.
The length of the array won't exceed 10,000.
All the integers in the given input belong to the range: [-1e7, 1e7].

 1 class Solution {
 2     func findPairs(_ nums: [Int], _ k: Int) -> Int {
 3         var count:Int = 0
 4         if k < 0 {return count}
 5         var hm:[Int:Int] = [Int:Int]()
 6         let len:Int = nums.count
 7         for i in 0..<len
 8         {
 9             hm[nums[i]] = i
10         }
11         for i in 0..<len
12         {
13 //如果字典包含所请求键的值，则下标返回包含该键的现有值的可选值。否则，下标返回nil：
14             if let keys = hm[nums[i] + k] 
15             {
16                 if keys != i
17                 {
18                     count += 1
19 //使用下标语法通过nil为该键指定值来从字典中删除键值对
20 //使用该removeValue(forKey:)方法从字典中删除键值对。hm.removeValue(forKey:key)
21 //此方法删除键值对（如果存在）并返回已删除的值，nil如果不存在值则返回
22                     hm[nums[i] + k] = nil                    
23                 }
24             }
25         }
26         return count
27     }
28 }

44ms

 1 class Solution {
 2     func findPairs(_ nums: [Int], _ k: Int) -> Int {
 3         guard k >= 0 else {
 4             return 0
 5         }
 6         
 7         var dict = [Int: Int]()
 8         for num in nums {
 9             //dict[num] = dict[num] == nil ? 0 : dict[num]! + 1
10             dict[num, default: 0] += 1
11         }
12         
13         var res = 0
14         for (num, count) in dict {
15             if k == 0 {
16                 if count > 1 {
17                     res += 1
18                 } 
19             } else {
20                 if dict[num - k] != nil {
21                     res += 1
22                 }
23             }
24             
25         }
26         return res
27     }
28 }

48ma

 1 class Solution {
 2     func findPairs(_ nums: [Int], _ k: Int) -> Int {
 3         guard nums.count > 1 else { return 0 }
 4         guard k >= 0 else { return 0 }
 5 
 6         var res = 0
 7         var dict = [Int: Int]()
 8         for n in nums {
 9             if let val = dict[n] {
10                 dict[n] = val + 1
11             } else {
12                 dict[n] = 1
13             }
14         }
15         
16         if k == 0 {
17             return dict.reduce(0) { (r, arg1) -> Int in
18                 if arg1.value > 1 {  return r + 1}
19                 return r
20             }
21         } else {
22             for key in dict.keys {
23                 if dict.keys.contains(key + k) {
24                     res = res + 1
25                 }
26             }
27             return res
28         }
29     }
30 }

52ms

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

56ms

 1 class Solution {
 2     func findPairs(_ nums: [Int], _ k: Int) -> Int {
 3         var dict = [Int: Bool]()
 4         
 5         if k < 0 {
 6             return 0
 7         }
 8         
 9         var diff = k < 0 ? -1*k : k
10         var count = 0
11         
12         
13         
14         for num in nums {
15             if !dict.keys.contains(num) {
16                 dict[num] = false
17                 if diff == 0 { continue }
18             }
19             
20             if dict.keys.contains(num-diff) {
21                 if dict[num-diff] == false {
22                     dict[num-diff] = true
23                     count = count + 1
24                 }
25                 
26             } 
27             if dict.keys.contains(num+diff) {
28                 if dict[num] == false {
29                     dict[num] = true
30                     count = count + 1
31                 }
32             }
33         }
34         
35         return count
36     }
37 }

68ms

 1 class Solution {
 2     func findPairs(_ nums: [Int], _ k: Int) -> Int {
 3         
 4         guard k >= 0 else { return 0 }
 5         
 6         var dict = nums.reduce(into: [Int: Int]()) { dict, num in
 7             dict[num, default:0] += 1
 8         }
 9         
10         guard k > 0 else {
11             return dict.reduce(into: 0) { count, numAndCount in
12                 if numAndCount.1 > 1 {
13                     count += 1
14                 }
15             }
16         }
17         
18         var totalCount = 0
19         for (num, count) in dict {
20             if dict[num+k, default:0] > 0 {
21                 totalCount += 1
22             }
23             if dict[num-k, default: 0] > 0 {
24                 totalCount += 1
25             }
26             dict[num] = 0
27         }
28         
29         return totalCount
30     }
31 }

76ms

 1 class Solution {
 2     func findPairs(_ nums: [Int], _ k: Int) -> Int {
 3         
 4         guard k >= 0 else { return 0 }
 5         
 6         var dict = nums.reduce(into: [Int: Int]()) { dict, num in
 7             dict[num, default:0] += 1
 8         }
 9         
10         if k == 0 {
11             return dict.values.reduce(into: 0) { total, appearances in
12                 if appearances > 1 {
13                     total += 1
14                 }
15             }
16         }
17         
18         return dict.keys.reduce(into: 0) { total, num in
19             if dict[num+k, default:0] > 0 {
20                 total += 1
21             }
22             if dict[num-k, default: 0] > 0 {
23                 total += 1
24             }
25             dict[num] = 0
26         }
27         
28     }
29 }

96ms

 1 class Solution {
 2     func findPairs(_ nums: [Int], _ k: Int) -> Int {
 3         guard nums.count > 1 else { return 0}
 4         let sortedNums = nums.sorted()
 5         var behind = 0
 6         var forward = 1
 7         var counter = 0
 8 
 9         while forward < nums.count , behind < nums.count {
10             if behind >= forward {
11                 forward = behind + 1
12                 continue
13             }
14             
15             let diff = abs(sortedNums[forward] - sortedNums[behind])
16             
17             if diff < k {
18                 repeat { forward += 1} while forward < nums.count && sortedNums[forward] == sortedNums[forward-1]
19                 continue
20             }
21 
22             if diff == k {
23                 counter += 1
24                 repeat { forward += 1} while forward < nums.count && sortedNums[forward] == sortedNums[forward-1]
25                 repeat { behind += 1} while behind < nums.count && sortedNums[behind] == sortedNums[behind-1]
26                 continue
27             }
28 
29             if diff > k {
30                 repeat { behind += 1} while behind < nums.count && sortedNums[behind] == sortedNums[behind-1]
31                 continue
32             }
33         }
34         return counter
35     }
36 }

116ms

 1 class Solution {
 2     func findPairs(_ nums: [Int], _ k: Int) -> Int {
 3         let nums =  nums.sorted()
 4         var i = 0
 5         var j = 1
 6         var res = 0
 7         while  i < nums.count && j < nums.count  {
 8             if nums[j] - nums[i] == k {
 9                 res += 1
10                 i += 1
11                 j += 1
12                 while j < nums.count && nums[j] == nums[j-1]  {
13                     j += 1
14                 }
15             }else if nums[j] - nums[i] < k {
16                 j += 1
17             }else  {
18                 i += 1
19                 j = i < j ? j : i+1
20             }
21         }
22         return res
23     }
24 }

132ms

 1 class Solution {
 2     func findPairs(_ nums: [Int], _ k: Int) -> Int {
 3         guard nums.count > 1 else {
 4             return 0
 5         }
 6         var resultSet: [[Int]] = []
 7         var sortedNums = nums
 8         sortedNums.sort()
 9         var firstPointer = 0
10         var secondPointer = 1
11         while secondPointer < sortedNums.count {
12             // print("comparing : (sortedNums[firstPointer]) and (sortedNums[secondPointer])")
13             if abs(sortedNums[firstPointer] - sortedNums[secondPointer]) == k {
14                 var validList: [Int] = [sortedNums[firstPointer], sortedNums[secondPointer]]
15                 // if !resultSet.contains(where: {$0 == validList}) {
16                     resultSet.append(validList)
17                 // }
18                 firstPointer += 1
19             } else if (sortedNums[secondPointer] - sortedNums[firstPointer]) < k {
20                 secondPointer += 1
21             } else {
22                 firstPointer += 1
23             }
24             if firstPointer == secondPointer {
25                 secondPointer += 1
26             }
27         }
28         print("This is the result set : (resultSet)")
29         var removeIndex:[Int] = []
30         if resultSet.count > 1 {
31             for index in 1..<resultSet.count {
32                 // print("index : (index)")
33                 if resultSet[index] == resultSet[index-1] {
34                     removeIndex.append(index)
35                 }
36             }
37         }
38         while removeIndex.count != 0 {
39             // print("removing (removeIndex.last) as (removeIndex)")
40             resultSet.remove(at: removeIndex.last!)
41             let _ = removeIndex.popLast()
42         }
43         return resultSet.count
44     }
45 }

查看全文

相关阅读:
spring4之依赖注入的三种方式
 Hibernate之总结
 Hibernate之dynamic-update
ThinkPhp调用webservice
Robot Framework：Web自动化之-元素处理
 Robot Framework：Web自动化之-元素定位
 RobotFramework：python+robotframework+selenium2library测试环境部署说明文档
 Robot Framework：Httplibrary库
 URL备忘
 Windows：CMD命令备忘

原文地址：https://www.cnblogs.com/strengthen/p/9813030.html