原文地址:http://www.javashuo.com/article/p-yduroxyn-cx.html html
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.app
Note that it is the kth smallest element in the sorted order, not the kth distinct element.less
Example:spa
matrix = [ [ 1, 5, 9], [10, 11, 13], [12, 13, 15] ], k = 8, return 13.
Note:
You may assume k is always valid, 1 ≤ k ≤ n^2.code
給定一個 n x n 矩陣,其中每行和每列元素均按升序排序,找到矩陣中第k小的元素。
請注意,它是排序後的第k小元素,而不是第k個元素。htm
示例:blog
matrix = [ [ 1, 5, 9], [10, 11, 13], [12, 13, 15] ], k = 8, 返回 13。
說明:
你能夠假設 k 的值永遠是有效的, 1 ≤ k ≤ n^2。排序
320mselement
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 }
384msget
1 class Solution { 2 func kthSmallest(_ matrix: [[Int]], _ k: Int) -> Int { 3 var tagArr = [Int](); 4 var totalItems = [Int](); 5 for itmes in matrix { 6 for item in itmes { 7 totalItems.append(item); 8 } 9 } 10 totalItems.sort(); 11 var num = totalItems[k-1]; 12 return num; 13 } 14 }