[Swift]LeetCode378. 有序矩阵中第K小的元素 | Kth Smallest Element in a Sorted Matrix

原文地址：https://www.cnblogs.com/strengthen/p/10280475.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.

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: 
You may assume k is always valid, 1 ≤ k ≤ n^2.

---

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。
请注意，它是排序后的第k小元素，而不是第k个元素。

示例:

matrix = [
   [ 1,  5,  9],
   [10, 11, 13],
   [12, 13, 15]
],
k = 8,

返回 13。

说明: 
你可以假设 k 的值永远是有效的, 1 ≤ k ≤ n^2。

---

320ms

class Solution {
    func kthSmallest(_ matrix: [[Int]], _ k: Int) -> Int {
        var matrix = matrix
        var left:Int = matrix[0][0]
        var right:Int = matrix.last!.last!
        while (left < right)
        {
            var mid:Int = left + (right - left) / 2
            var cnt:Int = search_less_equal(&matrix, mid)
            if cnt < k
            {
                left = mid + 1
            }
            else
            {
                right = mid
            }
        }
        return left
    }
    
    func search_less_equal(_ matrix:inout [[Int]], _ target: Int) -> Int
    {
        var n:Int = matrix.count
        var i:Int = n - 1
        var j:Int = 0
        var res:Int = 0
        while(i >= 0 && j < n)
        {
            if matrix[i][j] <= target
            {
                res += i + 1
                j += 1
            }
            else
            {
                i -= 1
            }
        }
        return res
    }
}

---

384ms

class Solution {
    func kthSmallest(_ matrix: [[Int]], _ k: Int) -> Int {
        var tagArr = [Int]();
        var totalItems = [Int]();
        for itmes in matrix {
            for item in itmes {
                totalItems.append(item);
            }
        }
        totalItems.sort();
        var num = totalItems[k-1];
        return num;
    }
}