矩阵中的最大路径和
给定一个 N * M 的矩阵。找到矩阵中的最大路径和。最大路径是从第一行到最后一行的所有元素的总和，您只能向下或沿对角线向左或向右移动。您可以从第一行的任何元素开始。
例子：
Input : mat[][] = 10 10 2 0 20 4
1 0 0 30 2 5
0 10 4 0 2 0
1 0 2 20 0 4
Output : 74
The maximum sum path is 20-30-4-20.
Input : mat[][] = 1 2 3
9 8 7
4 5 6
Output : 17
The maximum sum path is 3-8-6.我们得到一个 N * M 的矩阵。要首先找到最大路径和，我们必须在矩阵的第一行找到最大值。将此值存储在 res 中。现在对于矩阵更新元素中的每个元素，其最大值可以包含在最大路径中。如果该值大于 res 则更新 res。在最后一个返回 res 中，它由最大路径总和值组成。
C++
// CPP program for finding max path in matrix
#include
#define N 4
#define M 6
using namespace std;
// To calculate max path in matrix
int findMaxPath(int mat[][M])
{
for (int i = 1; i < N; i++) {
for (int j = 0; j < M; j++) {
// When all paths are possible
if (j > 0 && j < M - 1)
mat[i][j] += max(mat[i - 1][j],
max(mat[i - 1][j - 1],
mat[i - 1][j + 1]));
// When diagonal right is not possible
else if (j > 0)
mat[i][j] += max(mat[i - 1][j],
mat[i - 1][j - 1]);
// When diagonal left is not possible
else if (j < M - 1)
mat[i][j] += max(mat[i - 1][j],
mat[i - 1][j + 1]);
// Store max path sum
}
}
int res = 0;
for (int j = 0; j < M; j++)
res = max(mat[N-1][j], res);
return res;
}
// Driver program to check findMaxPath
int main()
{
int mat1[N][M] = { { 10, 10, 2, 0, 20, 4 },
{ 1, 0, 0, 30, 2, 5 },
{ 0, 10, 4, 0, 2, 0 },
{ 1, 0, 2, 20, 0, 4 } };
cout << findMaxPath(mat1) << endl;
return 0;
} Java
// Java program for finding max path in matrix
import static java.lang.Math.max;
class GFG
{
public static int N = 4, M = 6;
// Function to calculate max path in matrix
static int findMaxPath(int mat[][])
{
// To find max val in first row
int res = -1;
for (int i = 0; i < M; i++)
res = max(res, mat[0][i]);
for (int i = 1; i < N; i++)
{
res = -1;
for (int j = 0; j < M; j++)
{
// When all paths are possible
if (j > 0 && j < M - 1)
mat[i][j] += max(mat[i - 1][j],
max(mat[i - 1][j - 1],
mat[i - 1][j + 1]));
// When diagonal right is not possible
else if (j > 0)
mat[i][j] += max(mat[i - 1][j],
mat[i - 1][j - 1]);
// When diagonal left is not possible
else if (j < M - 1)
mat[i][j] += max(mat[i - 1][j],
mat[i - 1][j + 1]);
// Store max path sum
res = max(mat[i][j], res);
}
}
return res;
}
// driver program
public static void main (String[] args)
{
int mat[][] = { { 10, 10, 2, 0, 20, 4 },
{ 1, 0, 0, 30, 2, 5 },
{ 0, 10, 4, 0, 2, 0 },
{ 1, 0, 2, 20, 0, 4 }
};
System.out.println(findMaxPath(mat));
}
}
// Contributed by Pramod KumarPython3
# Python 3 program for finding max path in matrix
# To calculate max path in matrix
def findMaxPath(mat):
for i in range(1, N):
res = -1
for j in range(M):
# When all paths are possible
if (j > 0 and j < M - 1):
mat[i][j] += max(mat[i - 1][j],
max(mat[i - 1][j - 1],
mat[i - 1][j + 1]))
# When diagonal right is not possible
else if (j > 0):
mat[i][j] += max(mat[i - 1][j],
mat[i - 1][j - 1])
# When diagonal left is not possible
else if (j < M - 1):
mat[i][j] += max(mat[i - 1][j],
mat[i - 1][j + 1])
# Store max path sum
res = max(mat[i][j], res)
return res
# Driver program to check findMaxPath
N=4
M=6
mat = ([[ 10, 10, 2, 0, 20, 4 ],
[ 1, 0, 0, 30, 2, 5 ],
[ 0, 10, 4, 0, 2, 0 ],
[ 1, 0, 2, 20, 0, 4 ]])
print(findMaxPath(mat))
# This code is contributed by Azkia Anam.C#
// C# program for finding
// max path in matrix
using System;
class GFG
{
static int N = 4, M = 6;
// find the max element
static int max(int a, int b)
{
if(a > b)
return a;
else
return b;
}
// Function to calculate
// max path in matrix
static int findMaxPath(int [,]mat)
{
// To find max val
// in first row
int res = -1;
for (int i = 0; i < M; i++)
res = max(res, mat[0, i]);
for (int i = 1; i < N; i++)
{
res = -1;
for (int j = 0; j < M; j++)
{
// When all paths are possible
if (j > 0 && j < M - 1)
mat[i, j] += max(mat[i - 1, j],
max(mat[i - 1, j - 1],
mat[i - 1, j + 1]));
// When diagonal right
// is not possible
else if (j > 0)
mat[i, j] += max(mat[i - 1, j],
mat[i - 1, j - 1]);
// When diagonal left
// is not possible
else if (j < M - 1)
mat[i, j] += max(mat[i - 1, j],
mat[i - 1, j + 1]);
// Store max path sum
res = max(mat[i, j], res);
}
}
return res;
}
// Driver code
static public void Main (String[] args)
{
int[,] mat = {{10, 10, 2, 0, 20, 4},
{1, 0, 0, 30, 2, 5},
{0, 10, 4, 0, 2, 0},
{1, 0, 2, 20, 0, 4}};
Console.WriteLine(findMaxPath(mat));
}
}
// This code is contributed
// by Arnab KunduPHP
0 && $j < ($M - 1))
$mat[$i][$j] += max($mat[$i - 1][$j],
max($mat[$i - 1][$j - 1],
$mat[$i - 1][$j + 1]));
// When diagonal right is
// not possible
else if ($j > 0)
$mat[$i][$j] += max($mat[$i - 1][$j],
$mat[$i - 1][$j - 1]);
// When diagonal left is
// not possible
else if ($j < ($M - 1))
$mat[$i][$j] += max($mat[$i - 1][$j],
$mat[$i - 1][$j + 1]);
// Store max path sum
}
}
$res = 0;
for ($j = 0; $j < $M; $j++)
$res = max($mat[$N - 1][$j], $res);
return $res;
}
// Driver Code
$mat1 = array( array( 10, 10, 2, 0, 20, 4 ),
array( 1, 0, 0, 30, 2, 5 ),
array( 0, 10, 4, 0, 2, 0 ),
array( 1, 0, 2, 20, 0, 4 ));
echo findMaxPath($mat1),"\n";
// This code is contributed by Sach_Code
?>Javascript
输出：
74时间复杂度： O(N*M)