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[Swift]LeetCode219. 存在重复元素 II | Contains Duplicate II

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Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the absolute difference between i and j is at most k.

Example 1:

Input: nums = [1,2,3,1], k = 3
Output: true

Example 2:

Input: nums = [1,0,1,1], k = 1
Output: true

Example 3:

Input: nums = [1,2,3,1,2,3], k = 2
Output: false


给定一个整数数组和一个整数 k,判断数组中是否存在两个不同的索引 i 和 j,使得 nums [i] = nums [j],并且 i 和 j 的差的绝对值最大为 k

示例 1:

输入: nums = [1,2,3,1], k= 3
输出: true

示例 2:

输入: nums = [1,0,1,1], k=1
输出: true

示例 3:

输入: nums = [1,2,3,1,2,3], k=2
输出: false

 1 class Solution {
 2     func containsNearbyDuplicate(_ nums: [Int], _ k: Int) -> Bool {
 3         //判断
 4         if nums == nil || nums.count < 2 || k < 1
 5         {
 6             return false
 7         }
 8         var map:[Int:Int] = [Int:Int]()
 9         for i in 0..<(nums.count)
10         {
11             if map.keys.contains(nums[i])
12             {
13                 var sub:Int = i - map[nums[i]]!
14                 if sub <= k
15                 {
16                     return true
17                 }
18                 else
19                 {
20                     map[nums[i]] = i
21                 }
22             }
23             else
24             {
25                 map[nums[i]] = i
26             }
27         }
28         return false
29     }
30 }

32ms

 1 class Solution {
 2     func containsNearbyDuplicate(_ nums: [Int], _ k: Int) -> Bool {
 3         var set = Set<Int>()
 4         for i in 0..<nums.count {
 5             if i > k { set.remove(nums[i - k - 1]) }
 6             if !set.insert(nums[i]).inserted { return true }
 7         }
 8         return false
 9     }
10 }

32ms

 1 class Solution {
 2     func containsNearbyDuplicate(_ nums: [Int], _ k: Int) -> Bool {
 3         var set = Set<Int>()
 4         
 5         for i in 0..<nums.count {
 6             if i > k {
 7                 set.remove(nums[i - k - 1])
 8             }
 9             if !set.insert(nums[i]).inserted {
10                 return true
11             }
12         }
13         
14         return false
15     }
16 }

40ms

 1 class Solution {
 2     func containsNearbyDuplicate(_ nums: [Int], _ k: Int) -> Bool {
 3         var map: [Int:Int] = [:]
 4         
 5         for i in 0..<nums.count {
 6             if let index = map[nums[i]], i - index <= k {
 7                 return true
 8             } else {
 9                 map[nums[i]] = i
10             }
11         }
12         
13         return false
14     }
15 }

44ms

 1 class Solution {
 2     func containsNearbyDuplicate(_ nums: [Int], _ k: Int) -> Bool {
 3         var set: Set<Int> = []
 4         for i in 0..<nums.count {
 5             if i > k {
 6                 set.remove(nums[i-k-1])//移掉无效数据
 7             }
 8             if !set.insert(nums[i]).inserted {
 9                 return true
10             }
11         }
12         return false
13     }
14 }

 

posted @ 2018-10-05 19:36  为敢技术  阅读(432)  评论(0编辑  收藏  举报