[Swift]LeetCode378. 有序矩阵中第K小的元素 | Kth Smallest Ele
发布时间:2020-12-14 05:05:43 所属栏目:百科 来源:网络整理
导读:Given a? n ?x? n ?matrix where each of the rows and columns are sorted in ascending order,find the kth smallest element in the matrix. Note that it is the kth smallest element in the sorted order,not the kth distinct element. Example: matr
Given a?n?x?n?matrix where each of the rows and columns are sorted in ascending order,find the kth smallest element in the matrix. Note that it is the kth smallest element in the sorted order,not the kth distinct element. Example: matrix = [ [ 1,5,9],[10,11,13],[12,13,15] ],k = 8,return 13.? Note:? 给定一个?n x n?矩阵,其中每行和每列元素均按升序排序,找到矩阵中第k小的元素。 示例: matrix = [ [ 1,返回 13。 说明:? 320ms 1 class Solution { 2 func kthSmallest(_ matrix: [[Int]],_ k: Int) -> Int { 3 var matrix = matrix 4 var left:Int = matrix[0][0] 5 var right:Int = matrix.last!.last! 6 while (left < right) 7 { 8 var mid:Int = left + (right - left) / 2 9 var cnt:Int = search_less_equal(&matrix,mid) 10 if cnt < k 11 { 12 left = mid + 1 13 } 14 else 15 { 16 right = mid 17 } 18 } 19 return left 20 } 21 22 func search_less_equal(_ matrix:inout [[Int]],_ target: Int) -> Int 23 { 24 var n:Int = matrix.count 25 var i:Int = n - 1 26 var j:Int = 0 27 var res:Int = 0 28 while(i >= 0 && j < n) 29 { 30 if matrix[i][j] <= target 31 { 32 res += i + 1 33 j += 1 34 } 35 else 36 { 37 i -= 1 38 } 39 } 40 return res 41 } 42 } 384ms 1 class Solution { 2 func kthSmallest(_ matrix: [[Int]],_ k: Int) -> Int { 3 4 var tagArr = [Int](); 5 var totalItems = [Int](); 6 for itmes in matrix { 7 for item in itmes { 8 totalItems.append(item); 9 } 10 } 11 totalItems.sort(); 12 var num = totalItems[k-1]; 13 return num; 14 } 15 } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |