查找最大数量为1的行

给定一个布尔2D数组，其中对每一行进行排序。查找最大为1s的行。

例子：

Input matrix
0 1 1 1
0 0 1 1
1 1 1 1  // this row has maximum 1s
0 0 0 0

Output: 2

一种简单的方法是对矩阵进行逐行遍历，计算每行中1的数目，然后将计数与最大值进行比较。最后，返回最大为1s的行的索引。此方法的时间复杂度为O(m * n)，其中m是矩阵中的行数，n是矩阵中的列数。
我们可以做得更好。由于每一行都是已排序的，因此我们可以使用二进制搜索来计算每一行中的1。我们在每一行中找到1的第一个实例的索引。 1的计数等于列总数减去前1的索引。

请参见以下代码，以实现上述方法。

C++

// CPP program to find the row
// with maximum number of 1s
#include 
using namespace std;
#define R 4
#define C 4
 
// Function to find the index of first index
// of 1 in a boolean array arr[]
int first(bool arr[], int low, int high)
{
    if(high >= low)
    {
        // Get the middle index
        int mid = low + (high - low)/2;
     
        // Check if the element at middle index is first 1
        if ( ( mid == 0 || arr[mid-1] == 0) && arr[mid] == 1)
            return mid;
     
        // If the element is 0, recur for right side
        else if (arr[mid] == 0)
            return first(arr, (mid + 1), high);
         
        // If element is not first 1, recur for left side
        else
            return first(arr, low, (mid -1));
    }
    return -1;
}
 
// Function that returns index of row
// with maximum number of 1s.
int rowWithMax1s(bool mat[R][C])
{
    // Initialize max values
    int max_row_index = 0, max = -1;
 
    // Traverse for each row and count number of 1s
    // by finding the index of first 1
    int i, index;
    for (i = 0; i < R; i++)
    {
        index = first (mat[i], 0, C-1);
        if (index != -1 && C-index > max)
        {
            max = C - index;
            max_row_index = i;
        }
    }
 
    return max_row_index;
}
 
// Driver Code
int main()
{
    bool mat[R][C] = { {0, 0, 0, 1},
                    {0, 1, 1, 1},
                    {1, 1, 1, 1},
                    {0, 0, 0, 0}};
 
    cout << "Index of row with maximum 1s is " << rowWithMax1s(mat);
 
    return 0;
}
 
// This is code is contributed by rathbhupendra

C

// C program to find the row
// with maximum number of 1s
#include 
#define R 4
#define C 4
 
// Function to find the index of first index
// of 1 in a boolean array arr[]
int first(bool arr[], int low, int high)
{
if(high >= low)
{
    // Get the middle index
    int mid = low + (high - low)/2;
 
    // Check if the element at middle index is first 1
    if ( ( mid == 0 || arr[mid-1] == 0) && arr[mid] == 1)
    return mid;
 
    // If the element is 0, recur for right side
    else if (arr[mid] == 0)
    return first(arr, (mid + 1), high);
     
    // If element is not first 1, recur for left side
    else
    return first(arr, low, (mid -1));
}
return -1;
}
 
// Function that returns index of row
// with maximum number of 1s.
int rowWithMax1s(bool mat[R][C])
{
    // Initialize max values
    int max_row_index = 0, max = -1;
 
    // Traverse for each row and count number of 1s
    // by finding the index of first 1
    int i, index;
    for (i = 0; i < R; i++)
    {
    index = first (mat[i], 0, C-1);
    if (index != -1 && C-index > max)
    {
        max = C - index;
        max_row_index = i;
    }
    }
 
    return max_row_index;
}
 
// Driver Code
int main()
{
    bool mat[R][C] = { {0, 0, 0, 1},
                       {0, 1, 1, 1},
                       {1, 1, 1, 1},
                       {0, 0, 0, 0}};
 
    printf("Index of row with maximum 1s is %d "
                                , rowWithMax1s(mat));
 
    return 0;
}

Java

// Java program to find the row
// with maximum number of 1s
import java.io.*;
 
class GFG {
    static int R = 4, C = 4;
    // Function to find the index of first index
    // of 1 in a boolean array arr[]
    static int first(int arr[], int low, int high)
    {
        if (high >= low) {
            // Get the middle index
            int mid = low + (high - low) / 2;
 
            // Check if the element at middle index is first 1
            if ((mid == 0 || (arr[mid - 1] == 0)) && arr[mid] == 1)
                return mid;
 
            // If the element is 0, recur for right side
            else if (arr[mid] == 0)
                return first(arr, (mid + 1), high);
                 
            // If element is not first 1, recur for left side
            else
                return first(arr, low, (mid - 1));
        }
        return -1;
    }
 
    // Function that returns index of row
    // with maximum number of 1s.
    static int rowWithMax1s(int mat[][])
    {
        // Initialize max values
        int max_row_index = 0, max = -1;
 
        // Traverse for each row and count number of
        // 1s by finding the index of first 1
        int i, index;
        for (i = 0; i < R; i++) {
            index = first(mat[i], 0, C - 1);
            if (index != -1 && C - index > max) {
                max = C - index;
                max_row_index = i;
            }
        }
 
        return max_row_index;
    }
    // Driver Code
    public static void main(String[] args)
    {
        int mat[][] = { { 0, 0, 0, 1 },
                        { 0, 1, 1, 1 },
                        { 1, 1, 1, 1 },
                        { 0, 0, 0, 0 } };
        System.out.println("Index of row with maximum 1s is "
                                            + rowWithMax1s(mat));
    }
}
 
// This code is contributed by 'Gitanjali'.

Python 3

# Python3 program to find the row
# with maximum number of 1s
 
# Function to find the index
# of first index of 1 in a
# boolean array arr[]
def first( arr, low, high):
    if high >= low:
         
        # Get the middle index
        mid = low + (high - low)//2
 
        # Check if the element at
        # middle index is first 1
        if (mid == 0 or arr[mid - 1] == 0) and arr[mid] == 1:
            return mid
 
        # If the element is 0,
        # recur for right side
        elif arr[mid] == 0:
            return first(arr, (mid + 1), high)
     
        # If element is not first 1,
        # recur for left side
        else:
            return first(arr, low, (mid - 1))
    return -1
 
# Function that returns
# index of row with maximum
# number of 1s.
def rowWithMax1s( mat):
     
    # Initialize max values
    R = len(mat)
    C = len(mat[0])
    max_row_index = 0
    max = -1
     
    # Traverse for each row and
    # count number of 1s by finding
    #  the index of first 1
    for i in range(0, R):
        index = first (mat[i], 0, C - 1)
        if index != -1 and C - index > max:
            max = C - index
            max_row_index = i
 
    return max_row_index
 
# Driver Code
mat = [[0, 0, 0, 1],
       [0, 1, 1, 1],
       [1, 1, 1, 1],
       [0, 0, 0, 0]]
print ("Index of row with maximum 1s is",
      rowWithMax1s(mat))
 
# This code is contributed
# by shreyanshi_arun

C#

// C# program to find the row with maximum
// number of 1s
using System;
 
class GFG
{
public static int R = 4, C = 4;
 
// Function to find the index of first index
// of 1 in a boolean array arr[]
public static int first(int[] arr,
                        int low, int high)
{
    if (high >= low)
    {
        // Get the middle index
        int mid = low + (high - low) / 2;
 
        // Check if the element at middle
        // index is first 1
        if ((mid == 0 || (arr[mid - 1] == 0)) &&
                          arr[mid] == 1)
        {
            return mid;
        }
 
        // If the element is 0, recur
        // for right side
        else if (arr[mid] == 0)
        {
            return first(arr, (mid + 1), high);
        }
 
        // If element is not first 1, recur
        // for left side
        else
        {
            return first(arr, low, (mid - 1));
        }
    }
    return -1;
}
 
// Function that returns index of row
// with maximum number of 1s.
public static int rowWithMax1s(int[][] mat)
{
    // Initialize max values
    int max_row_index = 0, max = -1;
 
    // Traverse for each row and count number 
    // of 1s by finding the index of first 1
    int i, index;
    for (i = 0; i < R; i++)
    {
        index = first(mat[i], 0, C - 1);
        if (index != -1 && C - index > max)
        {
            max = C - index;
            max_row_index = i;
        }
    }
 
    return max_row_index;
}
 
// Driver Code
public static void Main(string[] args)
{
    int[][] mat = new int[][]
    {
        new int[] {0, 0, 0, 1},
        new int[] {0, 1, 1, 1},
        new int[] {1, 1, 1, 1},
        new int[] {0, 0, 0, 0}
    };
    Console.WriteLine("Index of row with maximum 1s is " +
                                       rowWithMax1s(mat));
}
}
 
// This code is contributed by Shrikant13

PHP

= $low)
    {
        // Get the middle index
        $mid = $low + intval(($high - $low) / 2);
     
        // Check if the element at middle
        // index is first 1
        if (($mid == 0 || $arr[$mid - 1] == 0) &&
                          $arr[$mid] == 1)
        return $mid;
     
        // If the element is 0, recur for
        // right side
        else if ($arr[$mid] == 0)
        return first($arr, ($mid + 1), $high);
         
        // If element is not first 1, recur
        // for left side
        else
        return first($arr, $low, ($mid - 1));
    }
    return -1;
}
 
// Function that returns index of row
// with maximum number of 1s.
function rowWithMax1s($mat)
{
     
    // Initialize max values
    $max_row_index = 0;
    $max = -1;
 
    // Traverse for each row and count number
    // of 1s by finding the index of first 1
     
    for ($i = 0; $i < R; $i++)
    {
    $index = first ($mat[$i], 0, (C - 1));
    if ($index != -1 && (C - $index) > $max)
    {
        $max = C - $index;
        $max_row_index = $i;
    }
    }
 
    return $max_row_index;
}
 
// Driver Code
$mat = array(array(0, 0, 0, 1),
             array(0, 1, 1, 1),
             array(1, 1, 1, 1),
             array(0, 0, 0, 0));
 
echo "Index of row with maximum 1s is " .              
                      rowWithMax1s($mat);
 
// This code is contributed by rathbhupendra
?>

输出：

Index of row with maximum 1s is 2

时间复杂度： O(mLogn)，其中m是矩阵中的行数，n是矩阵中的列数。
上述解决方案可以进一步优化。我们首先检查该行是否比最大行数多1，而不是对每一行进行二进制搜索。如果该行具有更多的1，则仅在该行中计数1。另外，要连续计算1秒，我们不会在整行中进行二进制搜索，而是在last max的索引之前进行搜索。

以下是上述解决方案的优化版本。

C++

#include 
using namespace std;
 
// The main function that returns index
// of row with maximum number of 1s.
int rowWithMax1s(bool mat[R][C])
{
    int i, index;
 
    // Initialize max using values from first row.
    int max_row_index = 0;
    int max = first(mat[0], 0, C - 1);
 
    // Traverse for each row and count number of 1s
    // by finding the index of first 1
    for (i = 1; i < R; i++)
    {
        // Count 1s in this row only if this row
        // has more 1s than max so far
 
        // Count 1s in this row only if this row
        // has more 1s than max so far
        if (max != -1 && mat[i][C - max - 1] == 1)
        {
            // Note the optimization here also
            index = first (mat[i], 0, C - max);
 
            if (index != -1 && C - index > max)
            {
                max = C - index;
                max_row_index = i;
            }
        }
        else
        {
            max = first(mat[i], 0, C - 1);
        }
    }
    return max_row_index;
}
 
// This code is contributed by rathbhupendra

C

// The main function that returns index of row with maximum number of 1s.
int rowWithMax1s(bool mat[R][C])
{
    int i, index;
  
    // Initialize max using values from first row. 
    int max_row_index = 0;
    int max = first(mat[0], 0, C-1);
  
    // Traverse for each row and count number of 1s by finding the index
    // of first 1
    for (i = 1; i < R; i++)
    {
        // Count 1s in this row only if this row has more 1s than
        // max so far
 
        // Count 1s in this row only if this row has more 1s than
        // max so far
        if (max != -1 && mat[i][C-max-1] == 1)
        {
            // Note the optimization here also
            index = first (mat[i], 0, C-max);
  
            if (index != -1 && C-index > max)
            {
                max = C - index;
                max_row_index = i;
            }  
        }
        else {
            max = first(mat[i], 0, C - 1);
        }  
    }  
    return max_row_index;
}

Java

public class gfg
{
  // The main function that returns index
  // of row with maximum number of 1s. 
  static int rowWithMax1s(boolean mat[][]) 
  { 
    int i, index; 
 
    // Initialize max using values from first row. 
    int max_row_index = 0; 
    int max = first(mat[0], 0, C - 1); 
 
    // Traverse for each row and count number of 1s 
    // by finding the index of first 1 
    for (i = 1; i < R; i++) 
    {
 
      // Count 1s in this row only if this row 
      // has more 1s than max so far 
 
      // Count 1s in this row only if this row 
      // has more 1s than max so far 
      if (max != -1 && mat[i][C - max - 1] == 1) 
      { 
        // Note the optimization here also 
        index = first (mat[i], 0, C - max); 
 
        if (index != -1 && C - index > max) 
        { 
          max = C - index; 
          max_row_index = i; 
        } 
      } 
      else
      { 
        max = first(mat[i], 0, C - 1); 
      } 
    } 
    return max_row_index; 
  }
}
 
// This code is contributed by divyesh072019.

Python3

# The main function that returns index
# of row with maximum number of 1s.
def rowWithMax1s(mat) : 
 
    # Initialize max using values from first row.
    max_row_index = 0;
    max = first(mat[0], 0, C - 1)
 
    # Traverse for each row and count number of 1s
    # by finding the index of first 1
    for i in range(1, R):
       
        # Count 1s in this row only if this row
        # has more 1s than max so far
 
        # Count 1s in this row only if this row
        # has more 1s than max so far
        if (max != -1 and mat[i][C - max - 1] == 1):
           
            # Note the optimization here also
            index = first (mat[i], 0, C - max)
 
            if (index != -1 and C - index > max):
                max = C - index
                max_row_index = i
        else:
            max = first(mat[i], 0, C - 1)
           
    return max_row_index;
 
# This code is contributed by Dharanendra L V

C＃

using System;
class GFG
{
     
    // The main function that returns index
    // of row with maximum number of 1s. 
    static int rowWithMax1s(bool[,] mat) 
    { 
        int i, index; 
       
        // Initialize max using values from first row. 
        int max_row_index = 0; 
        int max = first(mat[0], 0, C - 1); 
       
        // Traverse for each row and count number of 1s 
        // by finding the index of first 1 
        for (i = 1; i < R; i++) 
        {
           
            // Count 1s in this row only if this row 
            // has more 1s than max so far 
       
            // Count 1s in this row only if this row 
            // has more 1s than max so far 
            if (max != -1 && mat[i,C - max - 1] == 1) 
            { 
                // Note the optimization here also 
                index = first (mat[i], 0, C - max); 
       
                if (index != -1 && C - index > max) 
                { 
                    max = C - index; 
                    max_row_index = i; 
                } 
            } 
            else
            { 
                max = first(mat[i], 0, C - 1); 
            } 
        } 
        return max_row_index; 
    }
}
 
// This code is contributed by divyeshrabadiya07.

Java脚本

     
// The main function that returns index
// of row with maximum number of 1s.
function rowWithMax1s(mat)
{
    let i, index;
     
    // Initialize max using values from first row.
    let max_row_index = 0;
    let max = first(mat[0], 0, C - 1);
     
    // Traverse for each row and count number of 1s
    // by finding the index of first 1
    for(i = 1; i < R; i++)
    {
         
        // Count 1s in this row only if this row
        // has more 1s than max so far
         
        // Count 1s in this row only if this row
        // has more 1s than max so far
        if (max != -1 && mat[i][C - max - 1] == 1)
        {
             
            // Note the optimization here also
            index = first (mat[i], 0, C - max);
             
            if (index != -1 && C - index > max)
            {
                max = C - index;
                max_row_index = i;
            }
        }
        else
        {
            max = first(mat[i], 0, C - 1);
        }
    }
    return max_row_index;
}
 
// This code is contributed by suresh07
 

上述优化版本的最坏情况下的时间复杂度也是O(mLogn)，该遗嘱解决方案平均可以更好地工作。感谢Naveen Kumar Singh建议上述解决方案。
对于下面的矩阵，上述解决方案的最坏情况发生了。
0 0 0…0 1
0 0 0 ..0 1 1
0…0 1 1 1
….0 1 1 1 1

在最坏的情况下，以下方法适用于O(m + n)时间复杂度。
步骤1：获取第一行中第一个(或最左边的)1的索引。
步骤2：对第一行之后的每一行进行跟踪
…如果前最左1的左侧的元素为0，则忽略此行。
…ELSE向左移动，直到找到0。将最左边的索引更新为此索引，并将max_row_index更新为当前行。
时间复杂度为O(m + n)，因为我们可以尽可能快地走出第一步。

以下是此方法的实现。

C++

// The main function that returns index of row with maximum number of 1s.
int rowWithMax1s(bool mat[R][C])
{
    // Initialize first row as row with max 1s
    int max_row_index = 0;
 
    // The function first() returns index of first 1 in row 0.
    // Use this index to initialize the index of leftmost 1 seen so far
    int j = first(mat[0], 0, C-1);
    if (j == -1) // if 1 is not present in first row
      j = C - 1;
 
    for (int i = 1; i < R; i++)
    {
        // Move left until a 0 is found
        while (j >= 0 && mat[i][j] == 1)
        {
           j = j-1;  // Update the index of leftmost 1 seen so far
           max_row_index = i;  // Update max_row_index
        }
    }
    return max_row_index;
}

Java

// The main function that returns index of row with maximum number of 1s.
static int rowWithMax1s(boolean mat[][])
{
   
    // Initialize first row as row with max 1s
    int max_row_index = 0;
 
    // The function first() returns index of first 1 in row 0.
    // Use this index to initialize the index of leftmost 1 seen so far
    int j = first(mat[0], 0, C - 1);
    if (j == -1) // if 1 is not present in first row
      j = C - 1;
 
    for (int i = 1; i < R; i++)
    {
        // Move left until a 0 is found
        while (j >= 0 && mat[i][j] == 1)
        {
           j = j - 1;  // Update the index of leftmost 1 seen so far
           max_row_index = i;  // Update max_row_index
        }
    }
    return max_row_index;
}
 
 
// This code is contributed by rutvik_56.

C＃

// The main function that returns index of row with maximum number of 1s.
static int rowWithMax1s(bool[,] mat)
{
   
    // Initialize first row as row with max 1s
    int max_row_index = 0;
 
    // The function first() returns index of first 1 in row 0.
    // Use this index to initialize the index of leftmost 1 seen so far
    int j = first(mat[0], 0, C - 1);
    if (j == -1) // if 1 is not present in first row
      j = C - 1;
 
    for (int i = 1; i < R; i++)
    {
        // Move left until a 0 is found
        while (j >= 0 && mat[i, j] == 1)
        {
           j = j - 1;  // Update the index of leftmost 1 seen so far
           max_row_index = i;  // Update max_row_index
        }
    }
    return max_row_index;
}
 
 
// This code is contributed by Dharanendra L V.