给定一个数字的直角三角形，找出出现在从顶部到底部的路径上的最大数字总和，以便在每条路径上的下一个数字位于正下方或下方和一个位置-正确的。
例子 ：
Input : 1
1 2
4 1 2
2 3 1 1
Output : 9
Explanation : 1 + 1 + 4 + 3
Input : 2
4 1
1 2 7
Output : 10
Explanation : 2 + 1 + 7这个想法是找到在最后一行的每个单元格结束的最大和，并返回这些和的最大值。我们可以通过递归地考虑上述两个单元来递归地计算这些总和。由于存在重叠子问题，我们使用动态规划来找到在最后一行的特定单元格结束的最大和。
下面是上述想法的实现。
C++
// C++ program to print maximum sum
// in a right triangle of numbers
#include
using namespace std;
// function to find maximum sum path
int maxSum(int tri[][3], int n)
{
// Adding the element of row 1 to both the
// elements of row 2 to reduce a step from
// the loop
if (n > 1)
tri[1][1] = tri[1][1] + tri[0][0];
tri[1][0] = tri[1][0] + tri[0][0];
// Traverse remaining rows
for(int i = 2; i < n; i++) {
tri[i][0] = tri[i][0] + tri[i-1][0];
tri[i][i] = tri[i][i] + tri[i-1][i-1];
//Loop to traverse columns
for (int j = 1; j < i; j++){
// Checking the two conditions,
// directly below and below right.
// Considering the greater one
// tri[i] would store the possible
// combinations of sum of the paths
if (tri[i][j] + tri[i-1][j-1] >=
tri[i][j] + tri[i-1][j])
tri[i][j] = tri[i][j] + tri[i-1][j-1];
else
tri[i][j] = tri[i][j]+tri[i-1][j];
}
}
// array at n-1 index (tri[i]) stores
// all possible adding combination, finding
// the maximum one out of them
int max=tri[n-1][0];
for(int i=1;i Java
// Java program to print maximum sum
// in a right triangle of numbers
class GFG
{
// function to find maximum sum path
static int maxSum(int tri[][], int n)
{
// Adding the element of row 1 to both the
// elements of row 2 to reduce a step from
// the loop
if (n > 1)
tri[1][1] = tri[1][1] + tri[0][0];
tri[1][0] = tri[1][0] + tri[0][0];
// Traverse remaining rows
for(int i = 2; i < n; i++) {
tri[i][0] = tri[i][0] + tri[i-1][0];
tri[i][i] = tri[i][i] + tri[i-1][i-1];
//Loop to traverse columns
for (int j = 1; j < i; j++){
// Checking the two conditions,
// directly below and below right.
// Considering the greater one
// tri[i] would store the possible
// combinations of sum of the paths
if (tri[i][j] + tri[i-1][j-1] >=
tri[i][j] + tri[i-1][j])
tri[i][j] = tri[i][j]
+ tri[i-1][j-1];
else
tri[i][j] = tri[i][j]
+ tri[i-1][j];
}
}
// array at n-1 index (tri[i]) stores
// all possible adding combination,
// finding the maximum one out of them
int max = tri[n-1][0];
for(int i = 1; i < n; i++)
{
if(max < tri[n-1][i])
max = tri[n-1][i];
}
return max;
}
// Driver code
public static void main (String[] args)
{
int tri[][] = {{1}, {2,1}, {3,3,2}};
System.out.println(maxSum(tri, 3));
}
}
// This code is contributed by Anant Agarwal.Python
# Python program to print maximum sum
# in a right triangle of numbers.
# tri[][] is a 2D array that stores the
# triangle, n is number of lines or rows.
def maxSum(tri, n):
# Adding the element of row 1 to both the
# elements of row 2 to reduce a step from
# the loop
if n > 1:
tri[1][1] = tri[1][1]+tri[0][0]
tri[1][0] = tri[1][0]+tri[0][0]
# Traverse remaining rows
for i in range(2, n):
tri[i][0] = tri[i][0] + tri[i-1][0]
tri[i][i] = tri[i][i] + tri[i-1][i-1]
# Loop to traverse columns
for j in range(1, i):
# Checking the two conditions, directly below
# and below right. Considering the greater one
# tri[i] would store the possible combinations
# of sum of the paths
if tri[i][j]+tri[i-1][j-1] >= tri[i][j]+tri[i-1][j]:
tri[i][j] = tri[i][j] + tri[i-1][j-1]
else:
tri[i][j] = tri[i][j]+tri[i-1][j]
# array at n-1 index (tri[i]) stores all possible
# adding combination, finding the maximum one
# out of them
print max(tri[n-1])
# driver program
tri = [[1], [2,1], [3,3,2]]
maxSum(tri, 3)C#
// C# program to print
// maximum sum in a right
// triangle of numbers
using System;
class GFG
{
// function to find
// maximum sum path
static int maxSum(int [,]tri,
int n)
{
// Adding the element of row 1
// to both the elements of row 2
// to reduce a step from the loop
if (n > 1)
tri[1, 1] = tri[1, 1] +
tri[0, 0];
tri[1, 0] = tri[1, 0] +
tri[0, 0];
// Traverse remaining rows
for(int i = 2; i < n; i++)
{
tri[i, 0] = tri[i, 0] +
tri[i - 1, 0];
tri[i, i] = tri[i, i] +
tri[i - 1, i - 1];
//Loop to traverse columns
for (int j = 1; j < i; j++)
{
// Checking the two conditions,
// directly below and below right.
// Considering the greater one
// tri[i] would store the possible
// combinations of sum of the paths
if (tri[i, j] + tri[i - 1, j - 1] >=
tri[i, j] + tri[i - 1, j])
tri[i, j] = tri[i, j] +
tri[i - 1, j - 1];
else
tri[i, j] = tri[i, j] +
tri[i - 1, j];
}
}
// array at n-1 index (tri[i])
// stores all possible adding
// combination, finding the
// maximum one out of them
int max = tri[n - 1, 0];
for(int i = 1; i < n; i++)
{
if(max < tri[n - 1, i])
max = tri[n - 1, i];
}
return max;
}
// Driver Code
public static void Main ()
{
int [,]tri = {{1,0,0},
{2,1,0},
{3,3,2}};
Console.Write(maxSum(tri, 3));
}
}
// This code is contributed by ajit.PHP
1)
$tri[1][1] = $tri[1][1] + $tri[0][0];
$tri[1][0] = $tri[1][0] + $tri[0][0];
// Traverse remaining rows
for($i = 2; $i < $n; $i++)
{
$tri[$i][0] = $tri[$i][0] +
$tri[$i - 1][0];
$tri[$i][$i] = $tri[$i][$i] +
$tri[$i - 1][$i - 1];
//Loop to traverse columns
for ($j = 1; $j < $i; $j++)
{
// Checking the two conditions,
// directly below and below right.
// Considering the greater one
// tri[i] would store the possible
// combinations of sum of the paths
if ($tri[$i][$j] + $tri[$i - 1][$j - 1] >=
$tri[$i][$j] + $tri[$i - 1][$j])
$tri[$i][$j] = $tri[$i][$j] +
$tri[$i - 1][$j - 1];
else
$tri[$i][$j] = $tri[$i][$j] +
$tri[$i - 1][$j];
}
}
// array at n-1 index (tri[i])
// stores all possible adding
// combination, finding the
// maximum one out of them
$max = $tri[$n - 1][0];
for($i = 1; $i < $n; $i++)
{
if($max < $tri[$n - 1][$i])
$max = $tri[$n - 1][$i];
}
return $max;
}
// Driver Code
$tri = array(array(1),
array(2,1),
array(3,3,2));
echo maxSum($tri, 3);
// This code is contributed by ajit
?>Javascript
输出 ：
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