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)

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

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;
    }
}