排序矩阵中的搜索元素

给定排序的矩阵mat [n] [m]和元素’x’。找到x在矩阵中的位置(如果存在)，否则打印-1。矩阵的排序方式使得一行中的所有元素都以升序排序，并且对于行“ i”，其中1 <= i <= n-1，行“ i”的第一个元素大于或等于’i-1’行的最后一个元素。该方法应具有O(log n + log m)时间复杂度。
例子：

Input : mat[][] = { {1, 5, 9},
                    {14, 20, 21},
                    {30, 34, 43} }
        x = 14
Output : Found at (1, 0)

Input : mat[][] = { {1, 5, 9, 11},
                    {14, 20, 21, 26},
                    {30, 34, 43, 50} }
        x = 42
Output : -1

请注意，此问题不同于按行和按列排序的矩阵中的搜索。在这里，矩阵的排序更为严格，因为行的第一个元素大于前一行的最后一个元素。
一个简单的解决方案是将x与矩阵的每个元素一一比较。如果匹配，则返回位置。如果到达末尾，则返回-1。该解决方案的时间复杂度为O(nxm)。
一种有效的解决方案是将给定的2D数组类型转换为1D数组，然后对类型转换的数组应用二进制搜索。
下面说明了另一种不需要类型转换的有效方法。

1) Perform binary search on the middle column 
   till only two elements are left or till the
   middle element of some row in the search is
   the required element 'x'. This search is done
   to skip the rows that are not required
2) The two left elements must be adjacent. Consider
   the rows of two elements and do following
   a) check whether the element 'x' equals to the 
      middle element of any one of the 2 rows
   b) otherwise according to the value of the 
      element 'x' check whether it is present in 
      the 1st half of 1st row, 2nd half of 1st row, 
      1st half of 2nd row or 2nd half of 2nd row. 

Note: This approach works for the matrix n x m 
      where 2 <= n. The algorithm can be modified
      for matrix 1 x m, we just need to check whether
      2nd row exists or not      

例子：

Consider:    | 1  2  3  4| 
x = 3, mat = | 5  6  7  8|   Middle column:
             | 9 10 11 12|    = {2, 6, 10, 14}
             |13 14 15 16|   perform binary search on them
                             since, x < 6, discard the 
                             last 2 rows as 'a' will 
                             not lie in them(sorted matrix)
Now, only two rows are left
             | 1  2  3  4| 
x = 3, mat = | 5  6  7  8|   Check whether element is present
                             on the middle elements of these
                             rows = {2, 6}
                             x != 2 or 6
If not, consider the four sub-parts
1st half of 1st row = {1}, 2nd half of 1st row = {3, 4}
1st half of 2nd row = {5}, 2nd half of 2nd row = {7, 8}

According the value of 'x' it will be searched in the
2nd half of 1st row = {3, 4} and found at (i, j): (0, 2)                              

C++

// C++ implementation to search an element in a
// sorted matrix
#include 
using namespace std;
 
const int MAX = 100;
 
// This function does Binary search for x in i-th
// row. It does the search from mat[i][j_low] to
// mat[i][j_high]
void binarySearch(int mat[][MAX], int i, int j_low,
                                int j_high, int x)
{
    while (j_low <= j_high)
    {
        int j_mid = (j_low + j_high) / 2;
 
        // Element found
        if (mat[i][j_mid] == x)
        {
            cout << "Found at (" << i << ", "
                 << j_mid << ")";
            return;
        }
 
        else if (mat[i][j_mid] > x)
            j_high = j_mid - 1;
 
        else
            j_low = j_mid + 1;
    }
 
    // element not found
    cout << "Element no found";
}
 
// Function to perform binary search on the mid
// values of row to get the desired pair of rows
// where the element can be found
void sortedMatrixSearch(int mat[][MAX], int n,
                                  int m, int x)
{
    // Single row matrix
    if (n == 1)
    {
        binarySearch(mat, 0, 0, m-1, x);
        return;
    }
 
    // Do binary search in middle column.
    // Condition to terminate the loop when the
    // 2 desired rows are found
    int i_low = 0;
    int i_high = n-1;
    int j_mid = m/2;
    while ((i_low+1) < i_high)
    {
        int i_mid = (i_low + i_high) / 2;
 
        // element found
        if (mat[i_mid][j_mid] == x)
        {
            cout << "Found at (" << i_mid << ", "
                 << j_mid << ")";
            return;
        }
 
        else if (mat[i_mid][j_mid] > x)
            i_high = i_mid;
 
        else
            i_low = i_mid;
    }
 
    // If element is present on the mid of the
    // two rows
    if (mat[i_low][j_mid] == x)
        cout << "Found at (" << i_low << ","
             << j_mid << ")";
    else if (mat[i_low+1][j_mid] == x)
        cout << "Found at (" << (i_low+1)
             << ", " << j_mid << ")";
 
    // Ssearch element on 1st half of 1st row
    else if (x <= mat[i_low][j_mid-1])
        binarySearch(mat, i_low, 0, j_mid-1, x);
 
    // Search element on 2nd half of 1st row
    else if (x >= mat[i_low][j_mid+1]  &&
             x <= mat[i_low][m-1])
       binarySearch(mat, i_low, j_mid+1, m-1, x);
 
    // Search element on 1st half of 2nd row
    else if (x <= mat[i_low+1][j_mid-1])
        binarySearch(mat, i_low+1, 0, j_mid-1, x);
 
    // search element on 2nd half of 2nd row
    else
        binarySearch(mat, i_low+1, j_mid+1, m-1, x);
}
 
// Driver program to test above
int main()
{
    int n = 4, m = 5, x = 8;
    int mat[][MAX] = {{0, 6, 8, 9, 11},
                     {20, 22, 28, 29, 31},
                     {36, 38, 50, 61, 63},
                     {64, 66, 100, 122, 128}};
 
    sortedMatrixSearch(mat, n, m, x);
    return 0;
}

Java

// java implementation to search
// an element in a sorted matrix
import java.io.*;
 
class GFG
{
    static int MAX = 100;
     
    // This function does Binary search for x in i-th
    // row. It does the search from mat[i][j_low] to
    // mat[i][j_high]
    static void binarySearch(int mat[][], int i, int j_low,
                                    int j_high, int x)
    {
        while (j_low <= j_high)
        {
            int j_mid = (j_low + j_high) / 2;
     
            // Element found
            if (mat[i][j_mid] == x)
            {
                System.out.println ( "Found at (" + i
                                     + ", " + j_mid +")");
                return;
            }
     
            else if (mat[i][j_mid] > x)
                j_high = j_mid - 1;
     
            else
                j_low = j_mid + 1;
        }
     
        // element not found
        System.out.println ( "Element no found");
    }
     
    // Function to perform binary search on the mid
    // values of row to get the desired pair of rows
    // where the element can be found
    static void sortedMatrixSearch(int mat[][], int n,
                                         int m, int x)
    {
        // Single row matrix
        if (n == 1)
        {
            binarySearch(mat, 0, 0, m - 1, x);
            return;
        }
     
        // Do binary search in middle column.
        // Condition to terminate the loop when the
        // 2 desired rows are found
        int i_low = 0;
        int i_high = n - 1;
        int j_mid = m / 2;
        while ((i_low + 1) < i_high)
        {
            int i_mid = (i_low + i_high) / 2;
     
            // element found
            if (mat[i_mid][j_mid] == x)
            {
                System.out.println ( "Found at (" + i_mid +", "
                                    + j_mid +")");
                return;
            }
     
            else if (mat[i_mid][j_mid] > x)
                i_high = i_mid;
     
            else
                i_low = i_mid;
        }
     
        // If element is present on
        // the mid of the two rows
        if (mat[i_low][j_mid] == x)
            System.out.println ( "Found at (" + i_low + ","
                                 + j_mid +")");
        else if (mat[i_low + 1][j_mid] == x)
            System.out.println ( "Found at (" + (i_low + 1)
                                + ", " + j_mid +")");
     
        // Ssearch element on 1st half of 1st row
        else if (x <= mat[i_low][j_mid - 1])
            binarySearch(mat, i_low, 0, j_mid - 1, x);
     
        // Search element on 2nd half of 1st row
        else if (x >= mat[i_low][j_mid + 1] &&
                 x <= mat[i_low][m - 1])
        binarySearch(mat, i_low, j_mid + 1, m - 1, x);
     
        // Search element on 1st half of 2nd row
        else if (x <= mat[i_low + 1][j_mid - 1])
            binarySearch(mat, i_low + 1, 0, j_mid - 1, x);
     
        // search element on 2nd half of 2nd row
        else
            binarySearch(mat, i_low + 1, j_mid + 1, m - 1, x);
    }
     
    // Driver program
    public static void main (String[] args)
    {
        int n = 4, m = 5, x = 8;
        int mat[][] = {{0, 6, 8, 9, 11},
                       {20, 22, 28, 29, 31},
                       {36, 38, 50, 61, 63},
                       {64, 66, 100, 122, 128}};
     
        sortedMatrixSearch(mat, n, m, x);
         
    }
}
 
// This code is contributed by vt_m

Python3

# Python3 implementation
# to search an element in a
# sorted matrix
MAX = 100
 
# This function does Binary
# search for x in i-th
# row. It does the search
# from mat[i][j_low] to
# mat[i][j_high]
def binarySearch(mat, i, j_low,
                 j_high, x):
 
    while (j_low <= j_high):
     
        j_mid = (j_low + j_high) // 2
 
        # Element found
        if (mat[i][j_mid] == x):
         
            print("Found at (", i, ", ", j_mid, ")")
            return
 
        elif (mat[i][j_mid] > x):
            j_high = j_mid - 1
 
        else:
            j_low = j_mid + 1
    
    # Element not found
    print ("Element no found")
 
# Function to perform binary
# search on the mid values of
# row to get the desired pair of rows
# where the element can be found
def sortedMatrixSearch(mat, n, m, x):
 
    # Single row matrix
    if (n == 1):
     
        binarySearch(mat, 0, 0, m - 1, x)
        return
 
    # Do binary search in middle column.
    # Condition to terminate the loop
    # when the 2 desired rows are found
    i_low = 0
    i_high = n - 1
    j_mid = m // 2
    while ((i_low + 1) < i_high):
     
        i_mid = (i_low + i_high) // 2
 
        # element found
        if (mat[i_mid][j_mid] == x):
         
            print ("Found at (", i_mid, ", ", j_mid, ")")
            return
 
        elif (mat[i_mid][j_mid] > x):
            i_high = i_mid
 
        else:
            i_low = i_mid
 
    # If element is present on the mid of the
    # two rows
    if (mat[i_low][j_mid] == x):
        print ("Found at (" , i_low, ",", j_mid , ")")
    elif (mat[i_low + 1][j_mid] == x):
        print ("Found at (", (i_low + 1), ", ", j_mid, ")")
 
    # Ssearch element on 1st half of 1st row
    elif (x <= mat[i_low][j_mid - 1]):
        binarySearch(mat, i_low, 0, j_mid - 1, x)
 
    # Search element on 2nd half of 1st row
    elif (x >= mat[i_low][j_mid + 1] and
          x <= mat[i_low][m - 1]):
       binarySearch(mat, i_low, j_mid + 1, m - 1, x)
 
    # Search element on 1st half of 2nd row
    elif (x <= mat[i_low + 1][j_mid - 1]):
        binarySearch(mat, i_low + 1, 0, j_mid - 1, x)
  
    # Search element on 2nd half of 2nd row
    else:
        binarySearch(mat, i_low + 1, j_mid + 1, m - 1, x)
 
# Driver program to test above
if __name__ == "__main__":
 
    n = 4
    m = 5
    x = 8
    mat = [[0, 6, 8, 9, 11],
           [20, 22, 28, 29, 31],
           [36, 38, 50, 61, 63],
           [64, 66, 100, 122, 128]]
    sortedMatrixSearch(mat, n, m, x)
    
# This code is contributed by Chitranayal

C#

// C# implementation to search
// an element in a sorted matrix
using System;
 
class GFG
{
    // This function does Binary search for x in i-th
    // row. It does the search from mat[i][j_low] to
    // mat[i][j_high]
    static void binarySearch(int [,]mat, int i, int j_low,
                                        int j_high, int x)
    {
        while (j_low <= j_high)
        {
            int j_mid = (j_low + j_high) / 2;
     
            // Element found
            if (mat[i,j_mid] == x)
            {
                Console.Write ( "Found at (" + i +
                                ", " + j_mid +")");
                return;
            }
     
            else if (mat[i,j_mid] > x)
                j_high = j_mid - 1;
     
            else
                j_low = j_mid + 1;
        }
     
        // element not found
        Console.Write ( "Element no found");
    }
     
    // Function to perform binary search on the mid
    // values of row to get the desired pair of rows
    // where the element can be found
    static void sortedMatrixSearch(int [,]mat, int n,
                                        int m, int x)
    {
        // Single row matrix
        if (n == 1)
        {
            binarySearch(mat, 0, 0, m - 1, x);
            return;
        }
     
        // Do binary search in middle column.
        // Condition to terminate the loop when the
        // 2 desired rows are found
        int i_low = 0;
        int i_high = n - 1;
        int j_mid = m / 2;
        while ((i_low + 1) < i_high)
        {
            int i_mid = (i_low + i_high) / 2;
     
            // element found
            if (mat[i_mid,j_mid] == x)
            {
                 
                Console.Write ( "Found at (" + i_mid +
                                ", "    + j_mid +")");
                return;
            }
     
            else if (mat[i_mid,j_mid] > x)
                i_high = i_mid;
     
            else
                i_low = i_mid;
        }
     
        // If element is present on
        // the mid of the two rows
        if (mat[i_low,j_mid] == x)
        Console.Write ( "Found at (" + i_low +
                           "," + j_mid +")");
        else if (mat[i_low + 1,j_mid] == x)
        Console.Write ( "Found at (" + (i_low
                   + 1) + ", " + j_mid +")");
     
        // Ssearch element on 1st half of 1st row
        else if (x <= mat[i_low,j_mid - 1])
            binarySearch(mat, i_low, 0, j_mid - 1, x);
     
        // Search element on 2nd half of 1st row
        else if (x >= mat[i_low,j_mid + 1] &&
                 x <= mat[i_low,m - 1])
        binarySearch(mat, i_low, j_mid + 1, m - 1, x);
     
        // Search element on 1st half of 2nd row
        else if (x <= mat[i_low + 1,j_mid - 1])
            binarySearch(mat, i_low + 1, 0, j_mid - 1, x);
     
        // search element on 2nd half of 2nd row
        else
            binarySearch(mat, i_low + 1, j_mid + 1, m - 1, x);
    }
     
    // Driver program
    public static void Main (String[] args)
    {
        int n = 4, m = 5, x = 8;
        int [,]mat = {{0, 6, 8, 9, 11},
                    {20, 22, 28, 29, 31},
                    {36, 38, 50, 61, 63},
                    {64, 66, 100, 122, 128}};
     
        sortedMatrixSearch(mat, n, m, x);
    }
}
 
// This code is contributed by parashar...

输出

Found at (0,2)

时间复杂度：O(log n + log m)。找到两个所需的行需要O(Log n)时间。然后，在大小等于m / 2的四个部分之一中进行二进制搜索需要O(Log m)时间。

该方法由Ayush Jauhari贡献。

方法2：在二维中使用二进制搜索

该方法还具有相同的时间复杂度：O(log(m)+ log(n))和辅助空间：O(1)，但是该算法更加容易并且代码方式更易于理解。

方法：我们可以观察到，要查找的任何数字(例如k)必须存在于一行中，包括该行的第一个和最后一个元素(如果存在的话)。因此，我们首先使用二进制搜索(O(logn))找到k必须位于的行，然后再次使用二进制搜索在该行中搜索(O(logm))。

算法：

1)首先，我们将找到可能存在k = 2的正确行。为此，我们将在第一列和最后一列上同时应用二进制搜索。

    low=0, high=n-1

     i) if( k< first element of row(a[mid][0]) ) => k must exist in the row above

                                              => high=mid-1;

                                                    

     ii) if( k> last element of row(a[mid][m-1])) => k must exist in the row above

                                               => low=mid+1; 

                                                  

     iii) if( k> first element of row(a[mid][0]) &&  k< last element of row(a[mid][m-1]))

                                               => k must exist in this row

                                               => apply binary search in this row like in a 1-D array

     iv) i) if( k== first element of row(a[mid][0]) ||  k== last element of row(a[mid][m-1])) => found                                       

Example:
let k=2; n=3,m=4;
    matrix a: [0, 1, 2, 3 ]
              [10,11,12,13]
              [20,21,22,23]

1) low=0, high=n-1(=2) => mid=1 //check 1st row     [0....3]
                                                 -->[10...13]<-- 
                                                    [20...23]                                                        
   k < a[mid][0] => high = mid-1;(=1)
2) low=0, high=1; =>mid=0; //check 0th row    -->[0...3]<--  
    k>a[mid][0] && k k must exist in this row
    
    now simply apply binary search in 1-D array: [0,1,2,3]                              

以下是上述算法的Java实现：

Java

// Java program for the above approach
import java.util.*;
public class Main {
 
    static void findRow(int[][] a, int n, int m, int k)
    {
        int l = 0, r = n - 1, mid;
 
        while (l <= r) {
            mid = (l + r) / 2;
 
            // we'll check the left and
            // right most elements
            // of the row here itself
            // for efficiency
            if (k == a[mid][0]) // checking leftmost element
            {
                System.out.println("Found at (" + mid + ","
                                   + "0)");
                return;
            }
 
            if (k == a[mid][m - 1]) // checking rightmost
                                    // element
            {
                int t = m - 1;
                System.out.println("Found at (" + mid + ","
                                   + t + ")");
                return;
            }
 
            if (k > a[mid][0]
                && k < a[mid]
                        [m - 1]) // this means the element
                                 // must be within this row
            {
                binarySearch(a, n, m, k,
                             mid); // we'll apply binary
                                   // search on this row
                return;
            }
 
            if (k < a[mid][0])
                r = mid - 1;
            if (k > a[mid][m - 1])
                l = mid + 1;
        }
    }
 
    static void binarySearch(int[][] a, int n, int m, int k,
                             int x) // x is the row number
    {
        // now we simply have to apply binary search as we
        // did in a 1-D array, for the elements in row
        // number
        // x
 
        int l = 0, r = m - 1, mid;
        while (l <= r) {
            mid = (l + r) / 2;
 
            if (a[x][mid] == k) {
                System.out.println("Found at (" + x + ","
                                   + mid + ")");
                return;
            }
 
            if (a[x][mid] > k)
                r = mid - 1;
            if (a[x][mid] < k)
                l = mid + 1;
        }
        System.out.println("Element not found");
    }
   
    // Driver Code
    public static void main(String args[])
    {
        int n = 4; // no. of rows
        int m = 5; // no. of columns
 
        int a[][] = { { 0, 6, 8, 9, 11 },
                      { 20, 22, 28, 29, 31 },
                      { 36, 38, 50, 61, 63 },
                      { 64, 66, 100, 122, 128 } };
 
        int k = 31; // element to search
 
        findRow(a, n, m, k);
    }
}

Python3

# Python program for the above approach
def findRow(a, n, m, k):
    l = 0
    r = n - 1
    mid = 0
    while (l <= r) :
        mid = int((l + r) / 2)
         
        # we'll check the left and
        # right most elements
        # of the row here itself
        # for efficiency
        if(k == a[mid][0]): #checking leftmost element
            print("Found at (" , mid , ",", "0)", sep = "")
            return
         
        if(k == a[mid][m - 1]): # checking rightmost element
            t = m - 1
            print("Found at (" , mid , ",", t , ")", sep = "")
            return
        if(k > a[mid][0] and k < a[mid][m - 1]):    # this means the element
                                                    # must be within this row
            binarySearch(a, n, m, k, mid)    # we'll apply binary
                                            # search on this row
            return
        if (k < a[mid][0]):
            r = mid - 1
        if (k > a[mid][m - 1]):
            l = mid + 1
 
def binarySearch(a, n, m, k, x):    #x is the row number
     
    # now we simply have to apply binary search as we
    # did in a 1-D array, for the elements in row
    # number
    # x
    l = 0
    r = m - 1
    mid = 0
    while (l <= r):
        mid = int((l + r) / 2)
         
        if (a[x][mid] == k):
            print("Found at (" , x , ",", mid , ")", sep = "")
            return
        if (a[x][mid] > k):
            r = mid - 1
        if (a[x][mid] < k):
            l = mid + 1
     
    print("Element not found")
 
# Driver Code
n = 4 # no. of rows
m = 5 # no. of columns
a = [[ 0, 6, 8, 9, 11], [20, 22, 28, 29, 31], [36, 38, 50, 61, 63 ], [64, 66, 100, 122, 128]]
k = 31  # element to search
findRow(a, n, m, k)
 
# This code is contributed by avanitrachhadiya2155

C＃

// C# program for the above approach
using System;
public class GFG
{
 
  static void findRow(int[,] a, int n, int m, int k)
  {
    int l = 0, r = n - 1, mid;
 
    while (l <= r) {
      mid = (l + r) / 2;
 
      // we'll check the left and
      // right most elements
      // of the row here itself
      // for efficiency
      if (k == a[mid,0]) // checking leftmost element
      {
        Console.WriteLine("Found at (" + mid + ","
                          + "0)");
        return;
      }
 
      if (k == a[mid,m - 1]) // checking rightmost
        // element
      {
        int t = m - 1;
        Console.WriteLine("Found at (" + mid + ","
                          + t + ")");
        return;
      }
 
      if (k > a[mid,0]
          && k < a[mid,m - 1]) // this means the element
        // must be within this row
      {
        binarySearch(a, n, m, k,
                     mid); // we'll apply binary
        // search on this row
        return;
      }
 
      if (k < a[mid,0])
        r = mid - 1;
      if (k > a[mid,m - 1])
        l = mid + 1;
    }
  }
 
  static void binarySearch(int[,] a, int n, int m, int k,
                           int x) // x is the row number
  {
    // now we simply have to apply binary search as we
    // did in a 1-D array, for the elements in row
    // number
    // x
 
    int l = 0, r = m - 1, mid;
    while (l <= r) {
      mid = (l + r) / 2;
 
      if (a[x,mid] == k) {
        Console.WriteLine("Found at (" + x + ","
                          + mid + ")");
        return;
      }
 
      if (a[x,mid] > k)
        r = mid - 1;
      if (a[x,mid] < k)
        l = mid + 1;
    }
    Console.WriteLine("Element not found");
  }
 
  // Driver Code
  static public void Main ()
  {
    int n = 4; // no. of rows
    int m = 5; // no. of columns
 
    int[,] a = { { 0, 6, 8, 9, 11 },
                { 20, 22, 28, 29, 31 },
                { 36, 38, 50, 61, 63 },
                { 64, 66, 100, 122, 128 } };
 
    int k = 31; // element to search
 
    findRow(a, n, m, k);
  }
}
 
// This code is contributed by rag2127

输出

Found at (1,4)