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算法技巧:640. 求解方程的进阶指南

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在编程世界中,算法是解决问题的重要工具之一。本篇文章将带领大家深入解析LeetCode上的640. 求解方程问题,并使用C/C++编程语言提供详细的代码实现。

1. 题目

给定一个包含系数和变量的方程,求解方程并返回结果。

方程格式:Ax + B = C,其中:

  • A 是系数,可以为正数、负数或 0。
  • B 是系数,可以为正数、负数或 0。
  • C 是系数,可以为正数、负数或 0。
  • x 是变量。

2. 解题思路

解决该问题的方法是将方程两边同时减去 B,然后将等式两边同时除以 A,即可求得 x 的值。

3. 代码实现

#include <iostream>

using namespace std;

double solveEquation(string equation) {
    // Split the equation into left and right sides
    int equalIndex = equation.find('=');
    string left = equation.substr(0, equalIndex);
    string right = equation.substr(equalIndex + 1);

    // Parse the left side
    int leftCoefficient = 0;
    int leftConstant = 0;
    int leftIndex = 0;
    while (leftIndex < left.length()) {
        if (left[leftIndex] == 'x') {
            if (leftIndex == 0 || left[leftIndex - 1] == '+' || left[leftIndex - 1] == '-') {
                leftCoefficient += 1;
            } else {
                int coefficient = stoi(left.substr(leftIndex - 1, leftIndex));
                leftCoefficient += coefficient;
            }
        } else if (left[leftIndex] == '+' || left[leftIndex] == '-') {
            int constant = stoi(left.substr(leftIndex + 1));
            if (left[leftIndex] == '+') {
                leftConstant += constant;
            } else {
                leftConstant -= constant;
            }
        } else {
            int constant = stoi(left.substr(leftIndex));
            leftConstant += constant;
        }
        leftIndex++;
    }

    // Parse the right side
    int rightCoefficient = 0;
    int rightConstant = 0;
    int rightIndex = 0;
    while (rightIndex < right.length()) {
        if (right[rightIndex] == 'x') {
            if (rightIndex == 0 || right[rightIndex - 1] == '+' || right[rightIndex - 1] == '-') {
                rightCoefficient += 1;
            } else {
                int coefficient = stoi(right.substr(rightIndex - 1, rightIndex));
                rightCoefficient += coefficient;
            }
        } else if (right[rightIndex] == '+' || right[rightIndex] == '-') {
            int constant = stoi(right.substr(rightIndex + 1));
            if (right[rightIndex] == '+') {
                rightConstant += constant;
            } else {
                rightConstant -= constant;
            }
        } else {
            int constant = stoi(right.substr(rightIndex));
            rightConstant += constant;
        }
        rightIndex++;
    }

    // Calculate the value of x
    if (leftCoefficient == rightCoefficient) {
        if (leftConstant == rightConstant) {
            return -1; // Infinite solutions
        } else {
            return -1; // No solution
        }
    } else {
        double x = (rightConstant - leftConstant) / (leftCoefficient - rightCoefficient);
        return x;
    }
}

int main() {
    string equation;
    cout << "Enter the equation: ";
    cin >> equation;

    double result = solveEquation(equation);
    if (result == -1) {
        cout << "No solution or infinite solutions." << endl;
    } else {
        cout << "The solution is: " << result << endl;
    }

    return 0;
}

4. 总结

通过这篇文章,我们详细解析了LeetCode上的640. 求解方程问题,并提供了详细的C/C++代码实现。希望读者能够从中获得有益的知识。