给定一个包含数字和两个运算符“ +”和“ *”的表达式，我们需要找到可以通过使用不同的括号对该表达式求值而获得的最大值和最小值。
例子：
Input : expr = “1+2*3+4*5”
Output : Minimum Value = 27, Maximum Value = 105
Explanation:
Minimum evaluated value = 1 + (2*3) + (4*5) = 27
Maximum evaluated value = (1 + 2)*(3 + 4)*5 = 105我们可以通过动态编程的方法来解决这个问题，我们可以看到这个问题类似于矩阵链乘法，这里我们尝试使用不同的括号来最大化和最小化表达式的值，而不是矩阵乘法的数量。
在下面的代码中，我们首先将运算符和数字与给定的表达式分开，然后使用两个2D数组存储中间结果，这些结果类似于矩阵链乘法进行更新，并尝试根据数字之间的不同括号进行运算，但要根据它们之间的运算符。在最后一行，第一行的最后一个单元将最终结果存储在两个2D数组中。
CPP
// C++ program to get maximum and minimum
// values of an expression
#include
using namespace std;
// Utility method to check whether a character
// is operator or not
bool isOperator(char op)
{
return (op == '+' || op == '*');
}
// method prints minimum and maximum value
// obtainable from an expression
void printMinAndMaxValueOfExp(string exp)
{
vector num;
vector opr;
string tmp = "";
// store operator and numbers in different vectors
for (int i = 0; i < exp.length(); i++)
{
if (isOperator(exp[i]))
{
opr.push_back(exp[i]);
num.push_back(atoi(tmp.c_str()));
tmp = "";
}
else
{
tmp += exp[i];
}
}
// storing last number in vector
num.push_back(atoi(tmp.c_str()));
int len = num.size();
int minVal[len][len];
int maxVal[len][len];
// initializing minval and maxval 2D array
for (int i = 0; i < len; i++)
{
for (int j = 0; j < len; j++)
{
minVal[i][j] = INT_MAX;
maxVal[i][j] = 0;
// initializing main diagonal by num values
if (i == j)
minVal[i][j] = maxVal[i][j] = num[i];
}
}
// looping similar to matrix chain multiplication
// and updating both 2D arrays
for (int L = 2; L <= len; L++)
{
for (int i = 0; i < len - L + 1; i++)
{
int j = i + L - 1;
for (int k = i; k < j; k++)
{
int minTmp = 0, maxTmp = 0;
// if current operator is '+', updating tmp
// variable by addition
if(opr[k] == '+')
{
minTmp = minVal[i][k] + minVal[k + 1][j];
maxTmp = maxVal[i][k] + maxVal[k + 1][j];
}
// if current operator is '*', updating tmp
// variable by multiplication
else if(opr[k] == '*')
{
minTmp = minVal[i][k] * minVal[k + 1][j];
maxTmp = maxVal[i][k] * maxVal[k + 1][j];
}
// updating array values by tmp variables
if (minTmp < minVal[i][j])
minVal[i][j] = minTmp;
if (maxTmp > maxVal[i][j])
maxVal[i][j] = maxTmp;
}
}
}
// last element of first row will store the result
cout << "Minimum value : " << minVal[0][len - 1]
<< ", Maximum value : " << maxVal[0][len - 1];
}
// Driver code to test above methods
int main()
{
string expression = "1+2*3+4*5";
printMinAndMaxValueOfExp(expression);
return 0;
} Java
// Java program to get maximum and minimum
// values of an expression
import java.io.*;
import java.util.*;
class GFG
{
// Utility method to check whether a character
// is operator or not
static boolean isOperator(char op)
{
return (op == '+' || op == '*');
}
// method prints minimum and maximum value
// obtainable from an expression
static void printMinAndMaxValueOfExp(String exp)
{
Vector num = new Vector();
Vector opr = new Vector();
String tmp = "";
// store operator and numbers in different vectors
for (int i = 0; i < exp.length(); i++)
{
if (isOperator(exp.charAt(i)))
{
opr.add(exp.charAt(i));
num.add(Integer.parseInt(tmp));
tmp = "";
}
else
{
tmp += exp.charAt(i);
}
}
// storing last number in vector
num.add(Integer.parseInt(tmp));
int len = num.size();
int[][] minVal = new int[len][len];
int[][] maxVal = new int[len][len];
// initializing minval and maxval 2D array
for (int i = 0; i < len; i++)
{
for (int j = 0; j < len; j++)
{
minVal[i][j] = Integer.MAX_VALUE;
maxVal[i][j] = 0;
// initializing main diagonal by num values
if (i == j)
minVal[i][j] = maxVal[i][j]
= num.get(i);
}
}
// looping similar to matrix chain multiplication
// and updating both 2D arrays
for (int L = 2; L <= len; L++)
{
for (int i = 0; i < len - L + 1; i++)
{
int j = i + L - 1;
for (int k = i; k < j; k++)
{
int minTmp = 0, maxTmp = 0;
// if current operator is '+', updating
// tmp variable by addition
if (opr.get(k) == '+')
{
minTmp = minVal[i][k]
+ minVal[k + 1][j];
maxTmp = maxVal[i][k]
+ maxVal[k + 1][j];
}
// if current operator is '*', updating
// tmp variable by multiplication
else if (opr.get(k) == '*')
{
minTmp = minVal[i][k]
* minVal[k + 1][j];
maxTmp = maxVal[i][k]
* maxVal[k + 1][j];
}
// updating array values by tmp
// variables
if (minTmp < minVal[i][j])
minVal[i][j] = minTmp;
if (maxTmp > maxVal[i][j])
maxVal[i][j] = maxTmp;
}
}
}
// last element of first row will store the result
System.out.print(
"Minimum value : " + minVal[0][len - 1]
+ ", Maximum value : " + maxVal[0][len - 1]);
}
// Driver code to test above methods
public static void main(String[] args)
{
String expression = "1+2*3+4*5";
printMinAndMaxValueOfExp(expression);
}
}
// This code is contributed by Dharanendra L V. Python3
# Python3 program to get maximum and minimum
# values of an expression
# Utility method to check whether a character
# is operator or not
def isOperator(op):
return (op == '+' or op == '*')
# method prints minimum and maximum value
# obtainable from an expression
def printMinAndMaxValueOfExp(exp):
num = []
opr = []
tmp = ""
# store operator and numbers in different vectors
for i in range(len(exp)):
if (isOperator(exp[i])):
opr.append(exp[i])
num.append(int(tmp))
tmp = ""
else:
tmp += exp[i]
# storing last number in vector
num.append(int(tmp))
llen = len(num)
minVal = [[ 0 for i in range(llen)] for i in range(llen)]
maxVal = [[ 0 for i in range(llen)] for i in range(llen)]
# initializing minval and maxval 2D array
for i in range(llen):
for j in range(llen):
minVal[i][j] = 10**9
maxVal[i][j] = 0
# initializing main diagonal by num values
if (i == j):
minVal[i][j] = maxVal[i][j] = num[i]
# looping similar to matrix chain multiplication
# and updating both 2D arrays
for L in range(2, llen + 1):
for i in range(llen - L + 1):
j = i + L - 1
for k in range(i, j):
minTmp = 0
maxTmp = 0
# if current operator is '+', updating tmp
# variable by addition
if(opr[k] == '+'):
minTmp = minVal[i][k] + minVal[k + 1][j]
maxTmp = maxVal[i][k] + maxVal[k + 1][j]
# if current operator is '*', updating tmp
# variable by multiplication
elif(opr[k] == '*'):
minTmp = minVal[i][k] * minVal[k + 1][j]
maxTmp = maxVal[i][k] * maxVal[k + 1][j]
# updating array values by tmp variables
if (minTmp < minVal[i][j]):
minVal[i][j] = minTmp
if (maxTmp > maxVal[i][j]):
maxVal[i][j] = maxTmp
# last element of first row will store the result
print("Minimum value : ",minVal[0][llen - 1],", \
Maximum value : ",maxVal[0][llen - 1])
# Driver code
expression = "1+2*3+4*5"
printMinAndMaxValueOfExp(expression)
# This code is contributed by mohit kumar 29C#
// C# program to get maximum and minimum
// values of an expression
using System;
using System.Collections.Generic;
public class GFG
{
// Utility method to check whether a character
// is operator or not
static bool isOperator(char op)
{
return (op == '+' || op == '*');
}
// method prints minimum and maximum value
// obtainable from an expression
static void printMinAndMaxValueOfExp(string exp)
{
List num = new List();
List opr = new List();
string tmp = "";
// store operator and numbers in different vectors
for (int i = 0; i < exp.Length; i++)
{
if (isOperator(exp[i]))
{
opr.Add(exp[i]);
num.Add(int.Parse(tmp));
tmp = "";
}
else
{
tmp += exp[i];
}
}
// storing last number in vector
num.Add(int.Parse(tmp));
int len = num.Count;
int[,] minVal = new int[len,len];
int[,] maxVal = new int[len,len];
// initializing minval and maxval 2D array
for (int i = 0; i < len; i++)
{
for (int j = 0; j < len; j++)
{
minVal[i, j] = Int32.MaxValue;
maxVal[i, j] = 0;
// initializing main diagonal by num values
if (i == j)
{
minVal[i, j] = maxVal[i, j] = num[i];
}
}
}
// looping similar to matrix chain multiplication
// and updating both 2D arrays
for (int L = 2; L <= len; L++)
{
for (int i = 0; i < len - L + 1; i++)
{
int j = i + L - 1;
for (int k = i; k < j; k++)
{
int minTmp = 0, maxTmp = 0;
// if current operator is '+', updating
// tmp variable by addition
if (opr[k] == '+')
{
minTmp = minVal[i, k] + minVal[k + 1, j];
maxTmp = maxVal[i, k] + maxVal[k + 1, j];
}
// if current operator is '*', updating
// tmp variable by multiplication
else if (opr[k] == '*')
{
minTmp = minVal[i, k] * minVal[k + 1, j];
maxTmp = maxVal[i, k] * maxVal[k + 1, j];
}
// updating array values by tmp
// variables
if (minTmp < minVal[i, j])
minVal[i, j] = minTmp;
if (maxTmp > maxVal[i, j])
maxVal[i, j] = maxTmp;
}
}
}
// last element of first row will store the result
Console.Write("Minimum value : " +
minVal[0, len - 1] +
", Maximum value : " +
maxVal[0,len - 1]);
}
// Driver code to test above methods
static public void Main ()
{
string expression = "1+2*3+4*5";
printMinAndMaxValueOfExp(expression);
}
}
// This code is contributed by avanitrachhadiya2155 输出：
Minimum value : 27, Maximum value : 105