二进制矩阵中的最大十进制值路径

给定二进制平方矩阵[n * n]。在从左上角到右下角的路径中找到最大整数值。我们使用遍历路径的位计算整数值。我们从索引[0,0]开始，到索引[n-1] [n-1]结束。从索引[i，j]，我们可以移动[i，j + 1]或[i + 1，j]。

例子：

Input : mat[][] = {{1, 1, 0, 1},
                   {0, 1, 1, 0},
                   {1, 0, 0, 1},
                   {1, 0, 1, 1}}
Output : 111
Explanation : 
Path :   (0,0) -> (0,1) -> (1,1) -> (1,2) ->
         (2,2) -> (3,2) ->(3,3)  
Decimal value : 1*(2^0) + 1*(2^1) + 1*(2^2) + 1*(2^3) + 
                0*(2^4) + 1*(2^5) + 1*(2^6) = 111

可以按以下方式递归定义以上问题：

// p indicates power of 2, initially  p = i = j = 0
MaxDecimalValue(mat, i, j, p) 

   // If i or j is our of boundary
   If i >= n || j >= n  
      return 0

   // Compute rest of matrix find maximum decimal value 
   result  max(MaxDecimalValue(mat, i, j+1, p+1), 
               MaxDecimalValue(mat, i+1, j, p+1))

   If mat[i][j] == 1  
      return power(2, p) + result
   Else
      return result 

下面是上述递归算法的实现。

C++

// C++ program to find maximum decimal value path in
// binary matrix
#include
using namespace std;
 
#define N 4
 
// Returns maximum decimal value in binary matrix.
// Here p indicate power of 2
long long int maxDecimalValue(int mat[][N], int i, int j,
                                                   int p)
{
    // Out of matrix boundary
    if (i >= N || j >= N )
        return 0;
 
    int result = max(maxDecimalValue(mat, i, j+1, p+1),
                     maxDecimalValue(mat, i+1, j, p+1));
 
    // If current matrix value is 1 then return result +
    // power(2, p) else result
    if (mat[i][j] == 1)
        return pow(2, p) + result;
    else
        return result;
}
 
//Driver program
int main()
{
    int mat[][4] = {{ 1 ,1 ,0 ,1 },
        { 0 ,1 ,1 ,0 },
        { 1 ,0 ,0 ,1 },
        { 1 ,0 ,1 ,1 },
    };
 
    cout << maxDecimalValue(mat, 0, 0, 0) << endl;
    return 0;
}

Java

// Java program to find maximum decimal value path in
// binary matrix
 
class GFG {
 
    static final int N = 4;
 
// Returns maximum decimal value in binary matrix.
// Here p indicate power of 2
    static int maxDecimalValue(int mat[][], int i, int j,
            int p) {
        // Out of matrix boundary
        if (i >= N || j >= N) {
            return 0;
        }
 
        int result = Math.max(maxDecimalValue(mat, i, j + 1, p + 1),
                maxDecimalValue(mat, i + 1, j, p + 1));
 
        // If current matrix value is 1 then return result +
        // power(2, p) else result
        if (mat[i][j] == 1) {
            return (int) (Math.pow(2, p) + result);
        } else {
            return result;
        }
    }
 
// Driver program
    public static void main(String[] args) {
        int mat[][] = {{1, 1, 0, 1},
        {0, 1, 1, 0},
        {1, 0, 0, 1},
        {1, 0, 1, 1},};
 
        System.out.println(maxDecimalValue(mat, 0, 0, 0));
    }
}
//this code contributed by Rajput-Ji

Python3

# Python3 program to find maximum decimal
# value path in binary matrix
N =4
 
# Returns maximum decimal value in binary
# matrix. Here p indicate power of 2
def maxDecimalValue(mat, i, j, p):
 
    # Out of matrix boundary
    if i >= N or j >= N:
        return 0
         
    result = max(
        maxDecimalValue(mat, i, j+1, p+1),
        maxDecimalValue(mat, i+1, j, p+1))
 
    # If current matrix value is 1 then
    # return result + power(2, p) else
    # result
    if mat[i][j] == 1:
        return pow(2, p) + result
    else:
        return result
 
 
# Driver Program
mat = [ [1, 1, 0, 1],
        [0, 1, 1, 0],
        [1, 0, 0, 1],
        [1, 0, 1, 1] ]
 
print(maxDecimalValue(mat, 0, 0, 0))
 
# This code is contributed by Shrikant13.

C#

// C# program to find maximum decimal value path in
// binary matrix
 
using System;
class GFG {
 
    static int N = 4;
 
// Returns maximum decimal value in binary matrix.
// Here p indicate power of 2
    static int maxDecimalValue(int[,] mat, int i,
                                int j,int p)
    {
        // Out of matrix boundary
        if (i >= N || j >= N) {
            return 0;
        }
 
        int result = Math.Max(maxDecimalValue(mat, i, j + 1, p + 1),
                    maxDecimalValue(mat, i + 1, j, p + 1));
 
        // If current matrix value is 1 then return result +
        // power(2, p) else result
        if (mat[i,j] == 1)
        {
            return (int) (Math.Pow(2, p) + result);
        } else
        {
            return result;
        }
    }
 
// Driver program
    public static void Main() {
        int[,] mat = {{1, 1, 0, 1},
        {0, 1, 1, 0},
        {1, 0, 0, 1},
        {1, 0, 1, 1},};
 
        Console.Write(maxDecimalValue(mat, 0, 0, 0));
    }
}
// This code is contributed by Ita_c.

PHP

= $N || $j >= $N )
        return 0;
 
    $result = max(maxDecimalValue($mat, $i,
                            $j + 1, $p + 1),
                  maxDecimalValue($mat, $i + 1,
                                $j, $p + 1));
 
    // If current matrix value
    // is 1 then return result +
    // power(2, p) else result
    if ($mat[$i][$j] == 1)
        return pow(2, $p) + $result;
    else
        return $result;
}
 
    // Driver Code
    $mat = array(array(1 ,1 ,0 ,1),
                 array(0 ,1 ,1 ,0),
                 array(1 ,0 ,0 ,1),
                 array(1 ,0 ,1 ,1));
 
    echo maxDecimalValue($mat, 0, 0, 0) ;
     
// This code is contributed by nitin mittal.
?>

Javascript

 
// JavaScript program to find maximum
// decimal value path in binary matrix
let N = 4;
   
// Returns maximum decimal value in
// binary matrix.Here p indicate power of 2
function maxDecimalValue(mat, i, j, p)
{
     
    // Out of matrix boundary
    if (i >= N || j >= N)
    {
        return 0;
    }
 
    let result = Math.max(maxDecimalValue(mat, i, j + 1,
                                                  p + 1),
                          maxDecimalValue(mat, i + 1, j,
                                               p + 1));
                                                
    // If current matrix value is 1 then
    // return result + power(2, p) else result
    if (mat[i][j] == 1)
    {
        return (Math.pow(2, p) + result);
    }
    else
    {
        return result;
    }
}
 
// Driver Code
let mat = [ [ 1, 1, 0, 1 ],
            [ 0, 1, 1, 0 ],
            [ 1, 0, 0, 1 ],
            [ 1, 0, 1, 1 ] ];
 
document.write(maxDecimalValue(mat, 0, 0, 0));
 
// This code is contributed by souravghosh0416            
                          

---

输出：

111

上述递归解的时间复杂度是指数的。

Here matrix [3][3]               
            (2 2)
          /        \
     (1 2)          (2 1)
    /     \        /     \
 (0 2)   (1 1)   (1 1)   (2 1)
 /  \    /   \    /  \    / \ 
.    .  .    .    .   .   .  .
.    .  .    .    .   .   .  . and so no 

如果我们看到上述递归解的递归树，则可以观察到重叠的子问题。由于该问题具有重叠的子问题，因此我们可以使用动态编程有效地解决它。以下是基于动态编程的解决方案。

以下是使用动态编程实现上述问题的方法

C++

// C++ program to find Maximum decimal value Path in
// Binary matrix
#include
using namespace std;
#define N 4
 
// Returns maximum decimal value in binary matrix.
// Here p indicate power of 2
long long int MaximumDecimalValue(int mat[][N], int n)
{
    int dp[n][n];
    memset(dp, 0, sizeof(dp));
    if (mat[0][0] == 1)
        dp[0][0] = 1 ; // 1*(2^0)
 
    // Compute binary stream of first row of matrix
    // and store result in dp[0][i]
    for (int i=1; i

Java

// Java program to find Maximum decimal value Path in
// Binary matrix
public class GFG {
 
    final static int N = 4;
// Returns maximum decimal value in binary matrix.
// Here p indicate power of 2
 
    static int MaximumDecimalValue(int mat[][], int n) {
        int dp[][] = new int[n][n];
        if (mat[0][0] == 1) {
            dp[0][0] = 1; // 1*(2^0)
        }
        // Compute binary stream of first row of matrix
        // and store result in dp[0][i]
        for (int i = 1; i < n; i++) {
            // indicate 1*(2^i) + result of previous
            if (mat[0][i] == 1) {
                dp[0][i] = (int) (dp[0][i - 1] + Math.pow(2, i));
            } // indicate 0*(2^i) + result of previous
            else {
                dp[0][i] = dp[0][i - 1];
            }
        }
 
        // Compute binary stream of first column of matrix
        // and store result in dp[i][0]
        for (int i = 1; i < n; i++) {
            // indicate 1*(2^i) + result of previous
            if (mat[i][0] == 1) {
                dp[i][0] = (int) (dp[i - 1][0] + Math.pow(2, i));
            } // indicate 0*(2^i) + result of previous
            else {
                dp[i][0] = dp[i - 1][0];
            }
        }
 
        // Traversal rest Binary matrix and Compute maximum
        // decimal value
        for (int i = 1; i < n; i++) {
            for (int j = 1; j < n; j++) {
                // Here (i+j) indicate the current power of
                // 2 in path that is 2^(i+j)
                if (mat[i][j] == 1) {
                    dp[i][j] = (int) (Math.max(dp[i][j - 1], dp[i - 1][j])
                            + Math.pow(2, i + j));
                } else {
                    dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]);
                }
            }
        }
 
        // Return maximum decimal value in binary matrix
        return dp[n - 1][n - 1];
    }
 
// Driver program
    public static void main(String[] args) {
 
        int mat[][] = {{1, 1, 0, 1},
        {0, 1, 1, 0},
        {1, 0, 0, 1},
        {1, 0, 1, 1},};
        System.out.println(MaximumDecimalValue(mat, 4));
    }
}
/*This code is contributed by Rajput-Ji*/

Python3

# Python3 program to find Maximum decimal
# value Path in
# Binary matrix
 
N=4
 
# Returns maximum decimal value in binary matrix.
# Here p indicate power of 2
def MaximumDecimalValue(mat, n):
    dp=[[0 for i in range(n)] for i in range(n)]
    if (mat[0][0] == 1):
        dp[0][0] = 1     # 1*(2^0)
         
    # Compute binary stream of first row of matrix
    # and store result in dp[0][i]
    for i in range(1,n):
         
        # indicate 1*(2^i) + result of previous
        if (mat[0][i] == 1):
            dp[0][i] = dp[0][i-1] + 2**i
             
        # indicate 0*(2^i) + result of previous
        else:
            dp[0][i] = dp[0][i-1]
             
    # Compute binary stream of first column of matrix
    # and store result in dp[i][0]
    for i in range(1,n):
         
        # indicate 1*(2^i) + result of previous
        if (mat[i][0] == 1):
            dp[i][0] = dp[i-1][0] + 2**i
             
        # indicate 0*(2^i) + result of previous
    else:
        dp[i][0] = dp[i-1][0]
         
    # Traversal rest Binary matrix and Compute maximum
    # decimal value
    for i in range(1,n):
        for j in range(1,n):
             
            # Here (i+j) indicate the current power of
            # 2 in path that is 2^(i+j)
            if (mat[i][j] == 1):
                dp[i][j] = max(dp[i][j-1], dp[i-1][j])+(2**(i+j))
            else:
                dp[i][j] = max(dp[i][j-1], dp[i-1][j])
                 
    # Return maximum decimal value in binary matrix
    return dp[n-1][n-1]
 
# Driver program
if __name__=='__main__':
     
    mat = [[ 1 ,1 ,0 ,1 ],
          [ 0 ,1 ,1 ,0 ],
          [ 1 ,0 ,0 ,1 ],
          [ 1 ,0 ,1 ,1 ]]
           
    print (MaximumDecimalValue(mat, 4))
 
#this code is contributed by sahilshelangia

C＃

// C# program to find Maximum decimal value Path in
// Binary matrix
using System;
public class GFG {
  
    readonly static int N = 4;
// Returns maximum decimal value in binary matrix.
// Here p indicate power of 2
  
    static int MaximumDecimalValue(int [,]mat, int n) {
        int [,]dp = new int[n,n];
        if (mat[0,0] == 1) {
            dp[0,0] = 1; // 1*(2^0)
        }
        // Compute binary stream of first row of matrix
        // and store result in dp[0,i]
        for (int i = 1; i < n; i++) {
            // indicate 1*(2^i) + result of previous
            if (mat[0,i] == 1) {
                dp[0,i] = (int) (dp[0,i - 1] + Math.Pow(2, i));
            } // indicate 0*(2^i) + result of previous
            else {
                dp[0,i] = dp[0,i - 1];
            }
        }
  
        // Compute binary stream of first column of matrix
        // and store result in dp[i,0]
        for (int i = 1; i < n; i++) {
            // indicate 1*(2^i) + result of previous
            if (mat[i,0] == 1) {
                dp[i,0] = (int) (dp[i - 1,0] + Math.Pow(2, i));
            } // indicate 0*(2^i) + result of previous
            else {
                dp[i,0] = dp[i - 1,0];
            }
        }
  
        // Traversal rest Binary matrix and Compute maximum
        // decimal value
        for (int i = 1; i < n; i++) {
            for (int j = 1; j < n; j++) {
                // Here (i+j) indicate the current power of
                // 2 in path that is 2^(i+j)
                if (mat[i,j] == 1) {
                    dp[i,j] = (int) (Math.Max(dp[i,j - 1], dp[i - 1,j])
                            + Math.Pow(2, i + j));
                } else {
                    dp[i,j] = Math.Max(dp[i,j - 1], dp[i - 1,j]);
                }
            }
        }
  
        // Return maximum decimal value in binary matrix
        return dp[n - 1,n - 1];
    }
  
// Driver program
    public static void Main() {
  
        int [,]mat = {{1, 1, 0, 1},
        {0, 1, 1, 0},
        {1, 0, 0, 1},
        {1, 0, 1, 1},};
        Console.Write(MaximumDecimalValue(mat, 4));
    }
}
 
// This code is contributed by Rajput-Ji

Java脚本

// Javascript program to find Maximum decimal value Path in
// Binary matrix
     
    let N = 4;
    // Returns maximum decimal value in binary matrix.
    // Here p indicate power of 2
    function MaximumDecimalValue(mat,n)
    {
        let dp=new Array(n);
        for(let i = 0; i < n; i++)
        {
            dp[i] = new Array(n);
            for(let j = 0; j < n; j++)
            {
                dp[i][j] = 0;
            }
        }
         
        if (mat[0][0] == 1) {
            dp[0][0] = 1; // 1*(2^0)
        }
        // Compute binary stream of first row of matrix
        // and store result in dp[0][i]
        for (let i = 1; i < n; i++)
        {
         
            // indicate 1*(2^i) + result of previous
            if (mat[0][i] == 1)
            {
                dp[0][i] = dp[0][i - 1] + Math.pow(2, i);
            }
             
            // indicate 0*(2^i) + result of previous
            else
            {
                dp[0][i] = dp[0][i - 1];
            }
        }
  
        // Compute binary stream of first column of matrix
        // and store result in dp[i][0]
        for (let i = 1; i < n; i++)
        {
         
            // indicate 1*(2^i) + result of previous
            if (mat[i][0] == 1)
            {
                dp[i][0] = Math.floor(dp[i - 1][0] + Math.pow(2, i));
            }
             
            // indicate 0*(2^i) + result of previous
            else
            {
                dp[i][0] = dp[i - 1][0];
            }
        }
  
        // Traversal rest Binary matrix and Compute maximum
        // decimal value
        for (let i = 1; i < n; i++)
        {
            for (let j = 1; j < n; j++)
            {
             
                // Here (i+j) indicate the current power of
                // 2 in path that is 2^(i+j)
                if (mat[i][j] == 1)
                {
                    dp[i][j] = Math.floor(Math.max(dp[i][j - 1], dp[i - 1][j])
                            + Math.pow(2, i + j));
                }
                else
                {
                    dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]);
                }
            }
        }
  
        // Return maximum decimal value in binary matrix
        return dp[n - 1][n - 1];
    }
     
    // Driver program
    let mat = [[ 1 ,1 ,0 ,1 ],
          [ 0 ,1 ,1 ,0 ],
          [ 1 ,0 ,0 ,1 ],
          [ 1 ,0 ,1 ,1 ]];
    document.write(MaximumDecimalValue(mat, 4))
     
    // This code is contributed by rag2127.

输出：

111

时间复杂度： O(n ² )
辅助空间： O(n ² )