求解单变量线性方程

给定一个线性方程，任务是找到所用变量的值。该方程式仅包含“ +”，“-”运算，变量及其系数。

1. 如果方程无解，则返回“无解”。
2. 如果方程有无限解，则返回“无限解”。
3. 如果方程式只有一个解，请确保x的值为整数。

例子 ：

Input : "x + 5 - 3 + x = 6 + x - 2"
Output : "x = 2"

Input : "x = x"
Output : "Infinite solutions"

Input: "2x = x"
Output: "x = 0"

Input: "x = x + 2"
Output: "No solution"

方法：想法是使用两个指针来更新两个参数：使用的变量系数和总和。在“ =”的左侧和右侧，对每个数字使用相反的符号，由符号变量负责，一旦看到“ =”，该变量将翻转。

现在，在采用独特解决方案的情况下，有效总数与系数之比可以得出所需的结果。在无限解的情况下，有效总数和系数都为零，例如x + 1 = x +1。在没有解的情况下，x的系数结果为零，但有效总数为非零。

C++

// CPP program to solve the given equation
#include 
using namespace std;
  
// Function to solve the given equation
string solveEquation(string equation)
{
    int n = equation.size(), sign = 1, coeff = 0;
    int total = 0, i = 0;
  
    // Traverse the equation
    for (int j = 0; j < n; j++) {
        if (equation[j] == '+' || equation[j] == '-') {
            if (j > i)
                total += sign * stoi(equation.substr(i, j - i));
            i = j;
        }
  
        // For cases such as: x, -x, +x
        else if (equation[j] == 'x') {
            if ((i == j) || equation[j - 1] == '+')
                coeff += sign;
            else if (equation[j - 1] == '-')
                coeff -= sign;
            else
                coeff += sign * stoi(equation.substr(i, j - i));
            i = j + 1;
        }
  
        // Flip sign once '=' is seen
        else if (equation[j] == '=') {
            if (j > i)
                total += sign * stoi(equation.substr(i, j - i));
            sign = -1;
            i = j + 1;
        }
    }
  
    // There may be a number left in the end
    if (i < n)
        total += sign * stoi(equation.substr(i));
  
    // For infinite solutions
    if (coeff == 0 && total == 0)
        return "Infinite solutions";
  
    // For no solution
    if (coeff == 0 && total)
        return "No solution";
  
    // x = total sum / coeff of x
    // '-' sign indicates moving
    // numeric value to right hand side
    int ans = -total / coeff;
    return "x=" + to_string(ans);
}
  
// Driver code
int main()
{
    string equation = "x+5-3+x=6+x-2";
    cout << solveEquation(equation);
    return 0;
}

Java

// Java program to solve
// the given equation
import java.io.*;
  
class GFG
{
// Function to solve 
// the given equation
static String solveEquation(String equation)
{
    int n = equation.length(), 
        sign = 1, coeff = 0;
    int total = 0, i = 0;
  
    // Traverse the equation
    for (int j = 0; j < n; j++) 
    {
        if (equation.charAt(j) == '+' || 
            equation.charAt(j) == '-')
        {
            if (j > i)
                total += sign * 
                         Integer.parseInt(
                                 equation.substring(i, j));
            i = j;
        }
  
        // For cases such 
        // as: x, -x, +x
        else if (equation.charAt(j) == 'x')
        {
            if ((i == j) || 
                 equation.charAt(j - 1) == '+')
                coeff += sign;
                  
            else if (equation.charAt(j - 1) == '-')
                coeff -= sign;
                  
            else
                coeff += sign * 
                         Integer.parseInt(
                                 equation.substring(i, j));
            i = j + 1;
        }
  
        // Flip sign once 
        // '=' is seen
        else if (equation.charAt(j) == '=') 
        {
            if (j > i)
                total += sign * 
                         Integer.parseInt(
                                 equation.substring(i, j));
            sign = -1;
            i = j + 1;
        }
    }
  
    // There may be a 
    // number left in the end
    if (i < n)
        total = total + 
                sign * 
                Integer.parseInt(
                        equation.substring(i));
  
    // For infinite
    // solutions
    if (coeff == 0 && 
        total == 0)
        return "Infinite solutions";
  
    // For no solution
    if (coeff == 0 && 
        total != 0)
        return "No solution";
  
    // x = total sum / coeff 
    // of x '-' sign indicates 
    // moving numeric value to 
    // right hand side
    int ans = -total / coeff;
    return (Integer.toString(ans));
}
  
// Driver code
public static void main(String args[])
{
    String equation = new String("x+5-3+x=6+x-2");
    System.out.print("x = " + 
                      solveEquation(equation));
}
}
  
// This code is contributed by 
// Manish Shaw(manishshaw1)

Python3

# Python program to solve
# the given equation
  
# def to solve 
# the given equation
def solveEquation(equation) :
  
    n = len(equation) 
    sign = 1
    coeff = 0
    total = 0
    i = 0
  
    # Traverse the equation
    for j in range(0, n) :
      
        if (equation[j] == '+' or
            equation[j] == '-') :
          
            if (j > i) :
                total = (total + sign * 
                         int(equation[i: j]))
            i = j
          
        # For cases such 
        # as: x, -x, +x
        elif (equation[j] == 'x') :
          
            if ((i == j) or
                equation[j - 1] == '+') :
                coeff += sign
            elif (equation[j - 1] == '-') :
                coeff = coeff - sign
            else :
                coeff = (coeff + sign * 
                         int(equation[i: j]))
            i = j + 1
          
        # Flip sign once 
        # '=' is seen
        elif (equation[j] == '=') :
          
            if (j > i) :
                total = (total + sign * 
                         int(equation[i: j]))
            sign = -1
            i = j + 1
          
    # There may be a number
    # left in the end
    if (i < n) :
        total = (total + sign * 
                 int(equation[i: len(equation)]))
  
    # For infinite solutions
    if (coeff == 0 and
        total == 0) :
        return "Infinite solutions"
  
    # For no solution
    if (coeff == 0 and total) :
        return "No solution"
  
    # x = total sum / coeff of x
    # '-' sign indicates moving
    # numeric value to right hand side
    ans = -total / coeff
    return int(ans)
  
# Driver code
equation = "x+5-3+x=6+x-2"
print ("x = {}" . 
        format(solveEquation(equation)))
  
# This code is contributed by 
# Manish Shaw(manishshaw1)

C#

// C# program to solve
// the given equation
using System;
  
class GFG
{
    // Function to solve 
    // the given equation
    static string solveEquation(string equation)
    {
        int n = equation.Length, 
            sign = 1, coeff = 0;
        int total = 0, i = 0;
      
        // Traverse the equation
        for (int j = 0; j < n; j++) 
        {
            if (equation[j] == '+' || 
                equation[j] == '-')
            {
                if (j > i)
                    total += sign * 
                             Int32.Parse(
                             equation.Substring(i, j - i));
                i = j;
            }
      
            // For cases such 
            // as: x, -x, +x
            else if (equation[j] == 'x')
            {
                if ((i == j) || 
                     equation[j - 1] == '+')
                    coeff += sign;
                      
                else if (equation[j - 1] == '-')
                    coeff -= sign;
                      
                else
                    coeff += sign * 
                             Int32.Parse(
                             equation.Substring(i, j - i));
                i = j + 1;
            }
      
            // Flip sign once 
            // '=' is seen
            else if (equation[j] == '=') 
            {
                if (j > i)
                    total += sign * 
                             Int32.Parse(
                             equation.Substring(i, j - i));
                sign = -1;
                i = j + 1;
            }
        }
      
        // There may be a 
        // number left in the end
        if (i < n)
            total += sign * 
                     Int32.Parse(
                     equation.Substring(i));
      
        // For infinite
        // solutions
        if (coeff == 0 && total == 0)
            return "Infinite solutions";
      
        // For no solution
        if (coeff == 0 && total != 0)
            return "No solution";
      
        // x = total sum / coeff 
        // of x '-' sign indicates 
        // moving numeric value to 
        // right hand side
        int ans = -total / coeff;
        return "x = " + ans.ToString();
    }
      
    // Driver code
    static void Main()
    {
        string equation = "x+5-3+x=6+x-2";
        Console.Write(solveEquation(equation));
    }
}
  
// This code is contributed by 
// Manish Shaw(manishshaw1)

PHP

 $i)
                $total += $sign * 
                          intval(substr($equation, 
                                        $i, $j - $i));
            $i = $j;
        }
  
        // For cases such 
        // as: x, -x, +x
        else if ($equation[$j] == 'x') 
        {
            if (($i == $j) || 
                 $equation[$j - 1] == '+')
                $coeff += $sign;
            else if ($equation[$j - 1] == '-')
                $coeff -= $sign;
            else
                $coeff += $sign * 
                          intval(substr($equation, 
                                        $i, $j - $i));
            $i = $j + 1;
        }
  
        // Flip sign once 
        // '=' is seen
        else if ($equation[$j] == '=') 
        {
            if ($j > $i)
                $total += $sign * 
                          intval(substr($equation, 
                                        $i, $j - $i));
            $sign = -1;
            $i = $j + 1;
        }
    }
  
    // There may be a number
    // left in the end
    if ($i < $n)
        $total += $sign * 
                  intval(substr($equation, $i));
  
    // For infinite solutions
    if ($coeff == 0 && 
        $total == 0)
        return "Infinite solutions";
  
    // For no solution
    if ($coeff == 0 && $total)
        return "No solution";
  
    // x = total sum / coeff of x
    // '-' sign indicates moving
    // numeric value to right hand side
    $ans = -$total / $coeff;
    return "x = " . $ans;
}
  
// Driver code
$equation = "x+5-3+x=6+x-2";
echo (solveEquation($equation));
  
// This code is contributed by 
// Manish Shaw(manishshaw1)
?>

输出：

x = 2

时间复杂度： O(n)，其中n是等式字符串的长度。
辅助空间： O(1)