大佬教程收集整理的这篇文章主要介绍了[Swift]LeetCode315. 计算右侧小于当前元素的个数 | Count of Smaller Numbers After Self，大佬教程大佬觉得挺不错的，现在分享给大家，也给大家做个参考。
You are given an Integer array nums and you have to return a new countsarray. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i].
Example:
Input: [5,2,6,1] Output: To the right of 5 there are 2 smaller elements (2 and 1). To the right of 2 there is only 1 smaller element (1). To the right of 6 there is 1 smaller element (1). To the right of 1 there is 0 smaller element.[2,1,0] Explanation:
给定一个整数数组 nums，按要求返回一个新数组 counts。数组 counts 有该性质： counts[i] 的值是 nums[i] 右侧小于 nums[i] 的元素的数量。
示例:
输入: [5,1] 输出: 5 的右侧有 2 个更小的元素 (2 和 1). 2 的右侧仅有 1 个更小的元素 (1). 6 的右侧有 1 个更小的元素 (1). 1 的右侧有 0 个更小的元素.[2,0] 解释:
96ms
1 class Solution { 2 func countsmaller(_ nums: [Int]) -> [Int] { 3 var res: [Int] = [Int](repeaTing: 0,count: nums.count) 4 var vals = nums.sorted() 5 6 for i in 0..<nums.count { 7 8 let index = binarySearch(vals,nums[i]) 9 res[i] = index 10 vals.remove(at: indeX) 11 } 12 13 return res 14 } 15 16 func binarySearch(_ nums: [Int],_ val: int) -> Int { 17 18 var start = 0 19 var end = nums.count-1 20 var mid = (end - start) / 2 21 22 while start < end { 23 24 if nums[mid] >= val { 25 end = mid 26 } 27 else { 28 start = mid+1 29 } 30 mid = start + (end - start) / 2 31 } 32 33 return mid 34 } 35 }
272ms
1 class Solution { 2 func countsmaller(_ nums: [Int]) -> [Int] { 3 var res = [Int]() 4 5 var sorted = nums.sorted() 6 7 func inedxOf(_ v : Int,_ arr : [Int]) -> Int { 8 var l = 0 9 var r = arr.count-1 10 11 while l<=r { 12 let mid = (l+r) >> 1 13 if arr[mid] >= v { 14 r = mid - 1 15 }else { 16 l = mid + 1 17 } 18 } 19 20 return l 21 } 22 23 24 for i in 0..<nums.count { 25 let c = nums[i] 26 let index = inedxOf(c,sorted) 27 res.append(indeX) 28 sorted.remove(at: indeX) 29 } 30 31 return res 32 } 33 }
2692ms
1 class Solution { 2 func countsmaller(_ nums: [Int]) -> [Int] { 3 4 var counts = Array(repeaTing:0,count: nums.count) 5 6 for i in 0..<nums.count { 7 8 var nos = 0 9 10 for j in i+1..<nums.count { 11 12 if nums[j] < nums[i] { 13 nos += 1 14 } 15 } 16 17 counts[i] = nos 18 } 19 20 return counts 21 } 22 }
2924ms
1 class Solution { 2 func countsmaller(_ nums: [Int]) -> [Int] { 3 var results:[Int] = [] 4 for i in 0..<nums.count { 5 var smaller = 0 6 for j in i..<nums.count { 7 if nums[j] < nums[i] { 8 smaller += 1 9 } 10 } 11 results.append(smaller) 12 } 13 return results 14 } 15 }
4168ms
1 class Solution { 2 func countsmaller(_ nums: [Int]) -> [Int] { 3 var res = [Int].init() 4 guard nums.count>0 else { 5 return res 6 } 7 if nums.count == 1 { 8 return [0] 9 } 10 for i in 0..<nums.count-1 { 11 var count = 0 12 13 for j in (i+1)..<nums.count{ 14 if nums[j] < nums[i]{ 15 count += 1 16 } 17 } 18 res.append(count) 19 } 20 res.append(0) 21 return res 22 } 23 }
以上是大佬教程为你收集整理的[Swift]LeetCode315. 计算右侧小于当前元素的个数 | Count of Smaller Numbers After Self全部内容，希望文章能够帮你解决[Swift]LeetCode315. 计算右侧小于当前元素的个数 | Count of Smaller Numbers After Self所遇到的程序开发问题。
如果觉得大佬教程网站内容还不错，欢迎将大佬教程推荐给程序员好友。
本图文内容来源于网友网络收集整理提供，作为学习参考使用，版权属于原作者。
如您有任何意见或建议可联系处理。小编QQ：384754419，请注明来意。