从第 0 行的任何单元格开始到第 (N-1) 行的任何单元格结束的最大路径和

给定一个由正整数组成的 NXN 矩阵 Mat[N][N]。一个单元格只有三种可能的移动 (i, j)

1. (i+1, j)
2. (i+1, j-1)
3. (i+1, j+1)

从第 0 行的任何列开始，返回直到第 N-1 行的任何路径的最大总和。
例子：

Input : mat[4][4] = { {4, 2, 3, 4},
                      {2, 9, 1, 10},
                      {15, 1, 3, 0},
                      {16, 92, 41, 44} };
Output :120
path : 4 + 9 + 15 + 92 = 120 

提问：亚马逊面试

上述问题可以递归定义。
假设初始位置为 MaximumPathSum(N-1, j)，其中 j 从 0 到 N-1 变化。我们返回我们开始遍历的所有路径之间的最大值 (N-1, j) [其中 j 从 0 到 N-1]

i = N-1, j = 0 to N -1
int MaximumPath(Mat[][N], I, j)
 
  // IF we reached to first row of 
  // matrix then return value of that 
  // element  
  IF ( i == 0 && j = 0 )
  return    Mat[i][j]

  // out of matrix bound 
  IF( i = N || j < 0 )
   return 0;

  // call all rest position that we reached
  // from current position and find maximum 
  // between them and add current value in 
  // that path 
  return max(MaximumPath(Mat, i-1, j), 
             MaximumPath(Mat, i-1, j-1), 
             MaximumPath(Mat, i-1, j+1)))
             + Mat[i][j];

如果我们绘制上述递归解的递归树，我们可以观察到重叠的子问题。由于问题有重叠的子问题，我们可以使用动态规划有效地解决它。以下是基于动态规划的解决方案。

C++

// C++ program to find Maximum path sum
// start any column in row '0' and ends
// up to any column in row 'n-1'
#include
using namespace std;
#define N 4
 
// function find maximum sum path
int MaximumPath(int Mat[][N])
{
    int result = 0 ;
 
    // create 2D matrix to store the sum
    // of the path
    int dp[N][N+2];
 
    // initialize all dp matrix as '0'
    memset(dp, 0, sizeof(dp));
 
    // copy all element of first column into
    // 'dp' first column
    for (int i = 0 ; i < N ; i++)
        dp[0][i+1] = Mat[0][i];
 
    for (int i = 1 ; i < N ; i++)
        for (int j = 1 ; j <= N ; j++)
            dp[i][j] = max(dp[i-1][j-1],
                           max(dp[i-1][j],
                               dp[i-1][j+1])) +
                       Mat[i][j-1] ;
 
    // Find maximum path sum that end ups
    // at  any column of last row 'N-1'
    for (int i=0; i<=N; i++)
        result = max(result, dp[N-1][i]);
 
    // return maximum sum path
    return result ;
}
 
// driver program to test above function
int main()
{
    int Mat[4][4] = { { 4, 2 , 3 , 4 },
        { 2 , 9 , 1 , 10},
        { 15, 1 , 3 , 0 },
        { 16 ,92, 41, 44 }
    };
 
    cout << MaximumPath ( Mat ) <

---

Java

// Java program to find Maximum path sum
// start any column in row '0' and ends
// up to any column in row 'n-1'
import java.util.*;
 
class GFG {
     
    static int N = 4;
 
    // function find maximum sum path
    static int MaximumPath(int Mat[][])
    {
        int result = 0;
 
        // create 2D matrix to store the sum
        // of the path
        int dp[][] = new int[N][N + 2];
 
        // initialize all dp matrix as '0'
        for (int[] rows : dp)
            Arrays.fill(rows, 0);
 
        // copy all element of first column into
        // 'dp' first column
        for (int i = 0; i < N; i++)
            dp[0][i + 1] = Mat[0][i];
 
        for (int i = 1; i < N; i++)
            for (int j = 1; j <= N; j++)
                dp[i][j] = Math.max(dp[i - 1][j - 1],
                                    Math.max(dp[i - 1][j],
                                    dp[i - 1][j + 1])) +
                                    Mat[i][j - 1];
 
        // Find maximum path sum that end ups
        // at any column of last row 'N-1'
        for (int i = 0; i <= N; i++)
            result = Math.max(result, dp[N - 1][i]);
 
        // return maximum sum path
        return result;
    }
     
    // driver code
    public static void main(String arg[])
    {
        int Mat[][] = { { 4, 2, 3, 4 },
                        { 2, 9, 1, 10 },
                        { 15, 1, 3, 0 },
                        { 16, 92, 41, 44 } };
 
        System.out.println(MaximumPath(Mat));
    }
}
 
// This code is contributed by Anant Agarwal.

Python3

# Python3 program to find
# Maximum path sum
# start any column in
# row '0' and ends
# up to any column in row 'n-1'
 
N = 4
 
# function find maximum sum path
def MaximumPath(Mat):
 
    result = 0
 
    # create 2D matrix to store the sum
    # of the path
    # initialize all dp matrix as '0'
    dp = [[0 for i in range(N+2)] for j in range(N)]
 
    # copy all element of first column into
    # dp first column
    for i in range(N):
        for j in range(1, N+1):
            dp[i][j] = max(dp[i-1][j-1],
                           max(dp[i-1][j],
                               dp[i-1][j+1])) + \
                       Mat[i][j-1]
 
    # Find maximum path sum that end ups
    # at any column of last row 'N-1'
    for i in range(N+1):
        result = max(result, dp[N-1][i])
 
    # return maximum sum path
    return result
 
# driver program to test above function
Mat = [[4, 2, 3, 4],
       [2, 9, 1, 10],
       [15, 1, 3, 0],
       [16, 92, 41, 44]]
 
print(MaximumPath(Mat))
 
# This code is contributed by Soumen Ghosh.

C#

// C# program to find Maximum path sum
// start any column in row '0' and ends
// up to any column in row 'n-1'
using System;
 
class GFG {
     
    static int N = 4;
 
    // function find maximum sum path
    static int MaximumPath(int [,] Mat)
    {
        int result = 0;
 
        // create 2D matrix to store the sum
        // of the path
        int [,]dp = new int[N,N + 2];
 
        // initialize all dp matrix as '0'
        //for (int[] rows : dp)
        //    Arrays.fill(rows, 0);
 
        // copy all element of first column into
        // 'dp' first column
        for (int i = 0; i < N; i++)
            dp[0,i + 1] = Mat[0,i];
 
        for (int i = 1; i < N; i++)
            for (int j = 1; j <= N; j++)
                dp[i,j] = Math.Max(dp[i - 1,j - 1],
                                    Math.Max(dp[i - 1,j],
                                    dp[i - 1,j + 1])) +
                                    Mat[i,j - 1];
 
        // Find maximum path sum that end ups
        // at any column of last row 'N-1'
        for (int i = 0; i <= N; i++)
            result = Math.Max(result, dp[N - 1,i]);
 
        // return maximum sum path
        return result;
    }
     
    // driver code
    public static void Main()
    {
        int [,]Mat = { { 4, 2, 3, 4 },
                        { 2, 9, 1, 10 },
                        { 15, 1, 3, 0 },
                        { 16, 92, 41, 44 } };
 
        Console.WriteLine(MaximumPath(Mat));
    }
}
 
// This code is contributed by Ryuga.

Javascript

 
// Javascript program to find Maximum path sum
// start any column in row '0' and ends
// up to any column in row 'n-1'
     
    let N = 4;
     
    // function find maximum sum path
    function MaximumPath(Mat)
    {
        let result = 0;
   
        // create 2D matrix to store the sum
        // of the path
        let dp = new Array(N);
         
        // initialize all dp matrix as '0'
        for(let i=0;i<N;i++)
        {
            dp[i]=new Array(N+2);
            for(let j=0;j<N+2;j++)
            {
                dp[i][j]=0;
            }
        }
         
   
        // copy all element of first column into
        // 'dp' first column
        for (let i = 0; i < N; i++)
            dp[0][i + 1] = Mat[0][i];
   
        for (let i = 1; i < N; i++)
            for (let j = 1; j <= N; j++)
                dp[i][j] = Math.max(dp[i - 1][j - 1],
                                    Math.max(dp[i - 1][j],
                                    dp[i - 1][j + 1])) +
                                    Mat[i][j - 1];
   
        // Find maximum path sum that end ups
        // at any column of last row 'N-1'
        for (let i = 0; i <= N; i++)
            result = Math.max(result, dp[N - 1][i]);
   
        // return maximum sum path
        return result;
    }
     
    // driver code
    let Mat = [[4, 2, 3, 4],
       [2, 9, 1, 10],
       [15, 1, 3, 0],
       [16, 92, 41, 44]]
    document.write(MaximumPath(Mat))
      
    // This code is contributed by rag2127
     

输出：

120

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