搜索按行和按列排序的矩阵

给定一个nxn矩阵和一个数字x，找到x在矩阵中的位置(如果存在)。否则，打印“未找到”。在给定的矩阵中，每一行和每一列都按升序排序。设计的算法应具有线性时间复杂度。

例子：

Input: mat[4][4] = { {10, 20, 30, 40},
                      {15, 25, 35, 45},
                      {27, 29, 37, 48},
                      {32, 33, 39, 50}};
              x = 29
Output: Found at (2, 1)
Explanation: Element at (2,1) is 29

Input : mat[4][4] = { {10, 20, 30, 40},
                      {15, 25, 35, 45},
                      {27, 29, 37, 48},
                      {32, 33, 39, 50}};
              x = 100
Output : Element not found
Explanation: Element 100 is not found

简单的解决方案

• 方法：简单的想法是遍历数组并逐一搜索元素。
• 算法：
  1. 运行嵌套循环，外循环用于行，内循环用于列
  2. 用x检查每个元素，如果找到该元素，则打印“找到元素”
  3. 如果找不到元素，则打印“找不到元素”。
• 执行：

CPP14

// C++ program to search an element in row-wise
// and column-wise sorted matrix
#include 
  
using namespace std;
  
/* Searches the element x in mat[][]. If the
element is found, then prints its position
and returns true, otherwise prints "not found"
and returns false */
int search(int mat[4][4], int n, int x)
{
    if (n == 0)
        return -1;
     
    //traverse through the matrix
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < n; j++)
        //if the element is found
        if(mat[i][j] == x)
        {
            cout<<"Element found at ("<<
                        i << ", " << j << ")\n";
            return 1;
        }
    }
  
    cout << "n Element not found";
    return 0;
}
  
// Driver code
int main()
{
    int mat[4][4] = { { 10, 20, 30, 40 },
                      { 15, 25, 35, 45 },
                      { 27, 29, 37, 48 },
                      { 32, 33, 39, 50 } };
    search(mat, 4, 29);
  
    return 0;
}

Java

// Java program to search an element in row-wise
// and column-wise sorted matrix
 
class GFG {
  static int search(int[][] mat, int n, int x)
  {
    if (n == 0)
      return -1;
 
    // traverse through the matrix
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++)
        // if the element is found
        if (mat[i][j] == x) {
          System.out.print("Element found at ("
                           + i + ", " + j
                           + ")\n");
          return 1;
        }
    }
 
    System.out.print(" Element not found");
    return 0;
  }
  public static void main(String[] args)
  {
    int mat[][] = { { 10, 20, 30, 40 },
                   { 15, 25, 35, 45 },
                   { 27, 29, 37, 48 },
                   { 32, 33, 39, 50 } };
 
    search(mat, 4, 29);
  }
}

Python3

# Python program to search an element in row-wise
# and column-wise sorted matrix
 
# Searches the element x in mat[][]. If the
# element is found, then prints its position
# and returns true, otherwise prints "not found"
# and returns false
def search(mat, n, x):
    if(n == 0):
        return -1
     
    # Traverse through the matrix  
    for i in range(n):
        for j in range(n):
             
            # If the element is found
            if(mat[i][j] == x):
                print("Element found at (", i, ",", j, ")")
                return 1
     
    print(" Element not found")
    return 0
 
# Driver code
mat = [[10, 20, 30, 40], [15, 25, 35, 45],[27, 29, 37, 48],[32, 33, 39, 50]]
search(mat, 4, 29)
 
# This code is contributed by rag2127

C#

// C# program to search an element in row-wise
// and column-wise sorted matrix
using System;
 
class GFG{
 
/* Searches the element x in mat[][]. If the
element is found, then prints its position
and returns true, otherwise prints "not found"
and returns false */  
static int search(int[,] mat, int n, int x)
{
    if (n == 0)
        return -1;
     
    // Traverse through the matrix
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < n; j++)
         
        // If the element is found
        if (mat[i,j] == x)
        {
            Console.Write("Element found at (" + i +
                          ", " + j + ")\n");
            return 1;
        }
    }
    Console.Write(" Element not found");
    return 0;
}
 
// Driver code
static public void Main()
{
    int[,] mat = { { 10, 20, 30, 40 },
                   { 15, 25, 35, 45 },
                   { 27, 29, 37, 48 },
                   { 32, 33, 39, 50 } };
                    
    search(mat, 4, 29);
}
}
 
// This code is contributed by avanitrachhadiya2155

输出

Element found at (2, 1)

更好的解决方案是使用分而治之找到时间复杂度为O(n ^1.58 )的元素。详情请参阅这里。

高效的解决方案

• 方法：简单的想法是在每次比较中删除一行或一列，直到找到一个元素。从矩阵的右上角开始搜索。有三种可能的情况。
  1. 给定的数字大于当前的数字：这将确保当前行中的所有元素都小于给定的数字，因为指针已经在最右边的元素上并且对该行进行了排序。因此，整个行将被消除，并在下一行继续搜索。在这里消除意味着不需要搜索行。
  2. 给定的数字小于当前的数字：这将确保当前列中的所有元素都大于给定的数字。因此，整个列将被消除，并继续在前一列(即最左端的列)上进行搜索。
  3. 给定的号码等于当前号码：这将结束搜索。
• 算法：
  1. 假设给定元素为x，创建两个变量i = 0，j = n-1作为行和列的索引
  2. 循环运行直到i = 0
  3. 检查当前元素是否大于x，然后减少j的计数。排除当前列。
  4. 检查当前元素是否小于x，然后增加i的计数。排除当前行。
  5. 如果元素相等，则打印位置并结束。
• 感谢devendraiiit建议采用以下方法。
• 执行：

C++

// C++ program to search an element in row-wise
// and column-wise sorted matrix
#include 
 
using namespace std;
 
/* Searches the element x in mat[][]. If the
element is found, then prints its position
and returns true, otherwise prints "not found"
and returns false */
int search(int mat[4][4], int n, int x)
{
    if (n == 0)
        return -1;
   
    int smallest = mat[0][0], largest = mat[n - 1][n - 1];
    if (x < smallest || x > largest)
        return -1;
   
    // set indexes for top right element
    int i = 0, j = n - 1;
    while (i < n && j >= 0)
    {
        if (mat[i][j] == x)
        {
            cout << "n Found at "
                 << i << ", " << j;
            return 1;
        }
        if (mat[i][j] > x)
            j--;
       
        // Check if mat[i][j] < x
        else
            i++;
    }
 
    cout << "n Element not found";
    return 0;
}
 
// Driver code
int main()
{
    int mat[4][4] = { { 10, 20, 30, 40 },
                      { 15, 25, 35, 45 },
                      { 27, 29, 37, 48 },
                      { 32, 33, 39, 50 } };
    search(mat, 4, 29);
 
    return 0;
}
 
// This code is contributed
// by Akanksha Rai(Abby_akku)

C

// C program to search an element in row-wise
// and column-wise sorted matrix
#include 
 
/* Searches the element x in mat[][]. If the
element is found, then prints its position
and returns true, otherwise prints "not found"
and returns false */
int search(int mat[4][4], int n, int x)
{
    if (n == 0)
        return -1;
    int smallest = mat[0][0], largest = mat[n - 1][n - 1];
    if (x < smallest || x > largest)
        return -1;
   
    // set indexes for top right element
    int i = 0, j = n - 1;
    while (i < n && j >= 0)
    {
        if (mat[i][j] == x)
        {
            printf("\n Found at %d, %d", i, j);
            return 1;
        }
        if (mat[i][j] > x)
            j--;
        else // if mat[i][j] < x
            i++;
    }
 
    printf("n Element not found");
    return 0; // if ( i==n || j== -1 )
}
 
// driver program to test above function
int main()
{
    int mat[4][4] = {
        { 10, 20, 30, 40 },
        { 15, 25, 35, 45 },
        { 27, 29, 37, 48 },
        { 32, 33, 39, 50 },
    };
    search(mat, 4, 29);
    return 0;
}

Java

// JAVA Code for Search in a row wise and
// column wise sorted matrix
 
class GFG {
 
    /* Searches the element x in mat[][]. If the
    element is found, then prints its position
    and returns true, otherwise prints "not found"
    and returns false */
    private static void search(int[][] mat,
                                    int n, int x)
    {
         
        // set indexes for top right
        int i = 0, j = n - 1;
        // element
 
        while (i < n && j >= 0)
        {
            if (mat[i][j] == x)
            {
                System.out.print("n Found at " +
                                    i + " " + j);
                return;
            }
            if (mat[i][j] > x)
                j--;
            else // if mat[i][j] < x
                i++;
        }
 
        System.out.print("n Element not found");
        return; // if ( i==n || j== -1 )
    }
    // driver program to test above function
    public static void main(String[] args)
    {
        int mat[][] = { { 10, 20, 30, 40 },
                        { 15, 25, 35, 45 },
                        { 27, 29, 37, 48 },
                        { 32, 33, 39, 50 } };
 
        search(mat, 4, 29);
    }
}
// This code is contributed by Arnav Kr. Mandal.

Python3

# Python3 program to search an element
# in row-wise and column-wise sorted matrix
 
# Searches the element x in mat[][]. If the
# element is found, then prints its position
# and returns true, otherwise prints "not found"
# and returns false
def search(mat, n, x):
 
    i = 0
     
    # set indexes for top right element
    j = n - 1
    while ( i < n and j >= 0 ):
     
        if (mat[i][j] == x ):
     
            print("n Found at ", i, ", ", j)
            return 1
     
        if (mat[i][j] > x ):
            j -= 1
             
        # if mat[i][j] < x
        else:
            i += 1
     
    print("Element not found")
    return 0 # if (i == n || j == -1 )
 
# Driver Code
mat = [ [10, 20, 30, 40],
        [15, 25, 35, 45],
        [27, 29, 37, 48],
        [32, 33, 39, 50] ]
search(mat, 4, 29)
 
# This code is contributed by Anant Agarwal.

C＃

// C# Code for Search in a row wise and
// column wise sorted matrix
using System;
 
class GFG
{
    /* Searches the element x in mat[][]. If the
    element is found, then prints its position
    and returns true, otherwise prints "not found"
    and returns false */
    private static void search(int[, ] mat,
                               int n, int x)
    {
        // set indexes for top right
        // element
        int i = 0, j = n - 1;
 
        while (i < n && j >= 0)
        {
            if (mat[i, j] == x)
            {
                Console.Write("n Found at "
                              + i + ", " + j);
                return;
            }
 
            if (mat[i, j] > x)
                j--;
            else // if mat[i][j] < x
                i++;
        }
 
        Console.Write("n Element not found");
        return; // if ( i==n || j== -1 )
    }
    // driver program to test above function
    public static void Main()
    {
 
        int[, ] mat = { { 10, 20, 30, 40 },
                        { 15, 25, 35, 45 },
                        { 27, 29, 37, 48 },
                        { 32, 33, 39, 50 } };
 
        search(mat, 4, 29);
    }
}
 
// This code is contributed by Sam007

的PHP

= 0)
    {
        if ($mat[$i][$j] == $x)
        {
            echo "n found at " . $i.
                        ", " . $j;
            return 1;
        }
        if ($mat[$i][$j] > $x)
            $j--;
        else // if $mat[$i][$j] < $x
            $i++;
    }
     
    echo "n Element not found";
    return 0; // if ( $i==$n || $j== -1 )
}
 
// Driver Code
$mat = array(array(10, 20, 30, 40),
            array(15, 25, 35, 45),
            array(27, 29, 37, 48),
            array(32, 33, 39, 50));
search($mat, 4, 29);
 
// This code is contributed
// by ChitraNayal
?>

输出

n Found at 2, 1

时间复杂度： O(n)。
只需一个遍历，即i从0到n，j从n-1到0，最多2 * n步。
上述方法也适用于mxn矩阵(不仅适用于nxn)。复杂度为O(m + n)。
空间复杂度： O(1)。
不需要额外的空间。

https://youtu.be/_reu46LJtyk