给定大小为M * N的矩阵mat [] [] ，任务是找到所有矩阵元素，这些矩阵元素在其各自的行中最小，在其各自的列中最大。如果不存在这样的元素，则打印-1 。
例子：
Input: mat[][] = {{1, 10, 4}, {9, 3, 8}, {15, 16, 17}}
Output: 15
Explanation:
15 is the only element which is maximum in its column {1, 3, 15} and minimum in its row {15, 16, 17}.
Input: m[][] = {{10, 41}, {3, 5}, {16, 2}}
Output: -1
方法：请按照以下步骤解决问题：
- 创建一个unordered_set并存储矩阵每一行的最小元素。
- 遍历矩阵并找到每列的最大元素。对于每一列，检查获得的最大值是否已存在于unordered_set中。
- 如果发现是真的，请打印该号码。如果找不到这样的矩阵元素，则打印-1 。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Functionto find all the matrix elements
// which are minimum in its row and maximum
// in its column
vector minmaxNumbers(vector >& matrix,
vector& res)
{
// Initialize unordered set
unordered_set set;
// Traverse the matrix
for (int i = 0; i < matrix.size(); i++) {
int minr = INT_MAX;
for (int j = 0; j < matrix[i].size(); j++) {
// Update the minimum
// element of current row
minr = min(minr, matrix[i][j]);
}
// Insert the minimum
// element of the row
set.insert(minr);
}
for (int j = 0; j < matrix[0].size(); j++) {
int maxc = INT_MIN;
for (int i = 0; i < matrix.size(); i++) {
// Update the maximum
// element of current column
maxc = max(maxc, matrix[i][j]);
}
// Checking if it is already present
// in the unordered_set or not
if (set.find(maxc) != set.end()) {
res.push_back(maxc);
}
}
return res;
}
// Driver Code
int main()
{
vector > mat
= { { 1, 10, 4 },
{ 9, 3, 8 },
{ 15, 16, 17 } };
vector ans;
// Function call
minmaxNumbers(mat, ans);
// If no such matrix
// element is found
if (ans.size() == 0)
cout << "-1" << endl;
for (int i = 0; i < ans.size(); i++)
cout << ans[i] << endl;
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Functionto find all the matrix elements
// which are minimum in its row and maximum
// in its column
public static VectorminmaxNumbers(int[][] matrix,
Vector res)
{
// Initialize unordered set
Set set = new HashSet();
// Traverse the matrix
for(int i = 0; i < matrix.length; i++)
{
int minr = Integer.MAX_VALUE;
for(int j = 0; j < matrix[i].length; j++)
{
// Update the minimum
// element of current row
minr = Math.min(minr, matrix[i][j]);
}
// Insert the minimum
// element of the row
set.add(minr);
}
for(int j = 0; j < matrix[0].length; j++)
{
int maxc = Integer.MIN_VALUE;
for(int i = 0; i < matrix.length; i++)
{
// Update the maximum
// element of current column
maxc = Math.max(maxc, matrix[i][j]);
}
// Checking if it is already present
// in the unordered_set or not
if (set.contains(maxc))
{
res.add(maxc);
}
}
return res;
}
// Driver code
public static void main(String[] args)
{
int[][] mat = { { 1, 10, 4 },
{ 9, 3, 8 },
{ 15, 16, 17 } };
Vector ans = new Vector();
// Function call
ans = minmaxNumbers(mat, ans);
// If no such matrix
// element is found
if (ans.size() == 0)
System.out.println("-1");
for(int i = 0; i < ans.size(); i++)
System.out.println(ans.get(i));
}
}
// This code is contributed by divyeshrabadiya07 Python3
# Python3 program for the above approach
import sys
# Functionto find all the matrix elements
# which are minimum in its row and maximum
# in its column
def minmaxNumbers(matrix, res):
# Initialize unordered set
s = set()
# Traverse the matrix
for i in range(0, len(matrix), 1):
minr = sys.maxsize
for j in range(0, len(matrix[i]), 1):
# Update the minimum
# element of current row
minr = min(minr, matrix[i][j])
# Insert the minimum
# element of the row
s.add(minr)
for j in range(0, len(matrix[0]), 1):
maxc = -sys.maxsize - 1
for i in range(0, len(matrix), 1):
# Update the maximum
# element of current column
maxc = max(maxc, matrix[i][j])
# Checking if it is already present
# in the unordered_set or not
if (maxc in s):
res.append(maxc)
return res
# Driver Code
if __name__ == '__main__':
mat = [ [ 1, 10, 4 ],
[ 9, 3, 8 ],
[ 15, 16, 17 ] ]
ans = []
# Function call
minmaxNumbers(mat, ans)
# If no such matrix
# element is found
if (len(ans) == 0):
print("-1")
for i in range(len(ans)):
print(ans[i])
# This code is contributed by SURENDRA_GANGWARC#
// C# program for
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Functionto find all
// the matrix elements
// which are minimum
// in its row and
// maximum in its column
public static List minmaxNumbers(int[,] matrix,
List res)
{
// Initialize unordered set
HashSet set = new HashSet();
// Traverse the matrix
for(int i = 0; i < matrix.GetLength(0); i++)
{
int minr = int.MaxValue;
for(int j = 0; j < matrix.GetLength(1); j++)
{
// Update the minimum
// element of current row
minr = Math.Min(minr, matrix[i, j]);
}
// Insert the minimum
// element of the row
set.Add(minr);
}
for(int j = 0; j < matrix.GetLength(0); j++)
{
int maxc = int.MinValue;
for(int i = 0; i < matrix.GetLength(1); i++)
{
// Update the maximum
// element of current column
maxc = Math.Max(maxc, matrix[i, j]);
}
// Checking if it is already present
// in the unordered_set or not
if (set.Contains(maxc))
{
res.Add(maxc);
}
}
return res;
}
// Driver code
public static void Main(String[] args)
{
int[,] mat = {{1, 10, 4},
{9, 3, 8},
{15, 16, 17}};
List ans = new List();
// Function call
ans = minmaxNumbers(mat, ans);
// If no such matrix
// element is found
if (ans.Count == 0)
Console.WriteLine("-1");
for(int i = 0; i < ans.Count; i++)
Console.WriteLine(ans[i]);
}
}
// This code is contributed by 29AjayKumar 输出：
15
时间复杂度： O(M * N)
辅助空间： O(M + N)