How do I solve equations in the form 'x=5'?

Simply return the solution as 'x=5'. The equation is already balanced, so no further steps are necessary.

What if there is no 'x' in the equation?

If there is no 'x' and the constants do not balance, return 'No solution'. If the equation is always true, return 'Infinite solutions'.

What happens when the equation simplifies to 'x=x'?

This equation has infinite solutions. Return 'Infinite solutions' as the result.

Can the input equation contain terms with multiple variables?

No, this problem specifically deals with equations containing only one variable, 'x'.

How does GhostInterview help with solving this problem?

GhostInterview offers step-by-step assistance for parsing and balancing the equation, ensuring you handle all special cases, such as infinite or no solutions.

#640

Medium

auto_awesome数学·string

LeetCode 题解工作台

求解方程

求解一个给定的方程，将 x 以字符串 "x=#value" 的形式返回。该方程仅包含 '+' ， '-' 操作，变量 x 和其对应系数。如果方程没有解或存在的解不为整数，请返回 "No solution" 。如果方程有无限解，则返回 “Infinite solutions” 。题目保证，如果方程…

数学字符串模拟

题目描述

求解一个给定的方程，将x以字符串 "x=#value" 的形式返回。该方程仅包含 '+' ， '-' 操作，变量 x 和其对应系数。

如果方程没有解或存在的解不为整数，请返回 "No solution" 。如果方程有无限解，则返回 “Infinite solutions” 。

题目保证，如果方程中只有一个解，则 'x' 的值是一个整数。

示例 1：

输入: equation = "x+5-3+x=6+x-2"
输出: "x=2"

示例 2:

输入: equation = "x=x"
输出: "Infinite solutions"

示例 3:

输入: equation = "2x=x"
输出: "x=0"

提示:

3 <= equation.length <= 1000
equation 只有一个 '='.
方程由绝对值在 [0, 100] 范围内且无任何前导零的整数和变量 'x' 组成。

lightbulb

解题思路

方法一：数学

将方程 $equation$ 按照等号 “=” 切分为左右两个式子，分别算出左右两个式子中 "x" 的系数 $x_i$ ，以及常数的值 $y_i$ 。

那么方程转换为等式 $x_1 \times x + y_1 = x_2 \times x + y_2$ 。

当 $x_1 = x_2$ ：若 $y_1 \neq y_2$ ，方程无解；若 $y_1=y_2$ ，方程有无限解。
当 $x_1 \neq x_2$ ：方程有唯一解 $x=\frac{y_2-y_1}{x_1-x_2}$ 。

相似题目：

592. 分数加减运算

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class Solution:
    def solveEquation(self, equation: str) -> str:
        def f(s):
            x = y = 0
            if s[0] != '-':
                s = '+' + s
            i, n = 0, len(s)
            while i < n:
                sign = 1 if s[i] == '+' else -1
                i += 1
                j = i
                while j < n and s[j] not in '+-':
                    j += 1
                v = s[i:j]
                if v[-1] == 'x':
                    x += sign * (int(v[:-1]) if len(v) > 1 else 1)
                else:
                    y += sign * int(v)
                i = j
            return x, y

        a, b = equation.split('=')
        x1, y1 = f(a)
        x2, y2 = f(b)
        if x1 == x2:
            return 'Infinite solutions' if y1 == y2 else 'No solution'
        return f'x={(y2 - y1) // (x1 - x2)}'

speed

复杂度分析

指标	值
时间	and space complexity depend on how you process the equation. Parsing the string and performing arithmetic operations both involve linear time complexity, O(n), where n is the length of the equation. Space complexity depends on how the equation is stored and split, typically O(n).
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Look for candidates who can break down and parse the string effectively.
question_mark
Evaluate how well they handle edge cases, such as infinite solutions or no solutions.
question_mark
Assess if they can balance the equation systematically and solve for 'x' without skipping steps.

warning

常见陷阱

外企场景

error
Overcomplicating the parsing process, leading to errors when extracting coefficients or constants.
error
Failing to account for edge cases, such as when the equation simplifies to 'x=x' (infinite solutions).
error
Mismanaging signs during the balancing step, which could lead to incorrect values for 'x'.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Equations with only addition or subtraction operations.
arrow_right_alt
Handling more complex terms such as fractions or higher powers of 'x'.
arrow_right_alt
Expanding the problem to handle multiple variables or more complicated expressions.

help

常见问题

外企场景

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#592 分数加减运算 #537 复数乘法 #415 字符串相加