> For the complete documentation index, see [llms.txt](https://wentao-shao.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://wentao-shao.gitbook.io/leetcode/string/640.solve-the-equation.md).

# 640.Solve-the-Equation

## 640. Solve the Equation

## 题目地址

<https://leetcode.com/problems/solve-the-equation/>

## 题目描述

```
Solve a given equation and return the value of x in the form of string "x=#value". The equation contains only '+', '-' operation, the variable x and its coefficient.

If there is no solution for the equation, return "No solution".

If there are infinite solutions for the equation, return "Infinite solutions".

If there is exactly one solution for the equation, we ensure that the value of x is an integer.

Example 1:
Input: "x+5-3+x=6+x-2"
Output: "x=2"
Example 2:
Input: "x=x"
Output: "Infinite solutions"
Example 3:
Input: "2x=x"
Output: "x=0"
Example 4:
Input: "2x+3x-6x=x+2"
Output: "x=-1"
Example 5:
Input: "x=x+2"
Output: "No solution"
```

## 代码

### Approach 1: Partioning Coefficients

左边变量 = 右边常数

x = 右边/左边

* Time complexity : O(n)
* Space complexity : O(n)

```java
class Solution {
  public String solveEquation(String equation) {
        String[] lr = equation.split("=");
    int lhs = 0, rhs = 0;
    for (String x: breakIt(lr[0])) {
      if (x.indexOf("x") >= 0) {
        lhs += Integer.parseInt(coeff(x));
      } else {
        rhs -= Integer.parseInt(x);
      }
    }

      for (String x: breakIt(lr[1])) {
            if (x.indexOf("x") >= 0)
                lhs -= Integer.parseInt(coeff(x));
            else
                rhs += Integer.parseInt(x);
        }

    if (lhs == 0) {
      if (rhs == 0) {
        return "Infinite solutions";
      } else {
        return "No solution";
      }
    }
    return "x=" + rhs / lhs;
  }

  public List<String> breakIt(String s) {
    List<String> res = new ArrayList<>();
    String r = "";
    for (int i = 0; i < s.length(); i++) {
      if (s.charAt(i) == '+' || s.charAt(i) == '-') {
        if (r.length() > 0) { // 分割
          res.add(r);
        }
        r = "" + s.charAt(i);
      } else {
        r += s.charAt(i);
      }
    }
    res.add(r);
    return res;
  }

  public String coeff(String x) {
    if (x.length() > 1 && x.charAt(x.length() - 2) >= '0' && x.charAt(x.length() - 2) <= '9') {
      return x.replace("x", "");
    } else {
      return x.replace("x", "1");
    }
  }
}
```

### Approach #2 Using regex for spliting

```java
class Solution {
    public String coeff(String x) {
        if (x.length() > 1 && x.charAt(x.length() - 2) >= '0' && x.charAt(x.length() - 2) <= '9')
            return x.replace("x", "");
        return x.replace("x", "1");
    }
    public String solveEquation(String equation) {
        String[] lr = equation.split("=");
        int lhs = 0, rhs = 0;
        for (String x: lr[0].split("(?=\\+)|(?=-)")) {
            if (x.indexOf("x") >= 0) {
                lhs += Integer.parseInt(coeff(x));
            } else
                rhs -= Integer.parseInt(x);
        }
        for (String x: lr[1].split("(?=\\+)|(?=-)")) {
            if (x.indexOf("x") >= 0)
                lhs -= Integer.parseInt(coeff(x));
            else
                rhs += Integer.parseInt(x);
        }
        if (lhs == 0) {
            if (rhs == 0)
                return "Infinite solutions";
            else
                return "No solution";
        } else
            return "x=" + rhs / lhs;
    }
}
```