360笔试题-修复方程

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        for (int i = 0; i < T; i++) {
            String equa = sc.next();
            boolean res = canMakeEquationTrue(equa);
            if (res) {
                System.out.println("Yes");
            } else {
                System.out.println("No");
            }
        }
        sc.close();
    }

    private static boolean canMakeEquationTrue(String equa) {
        String[] parts = equa.split("=");
        String leftPart = parts[0];
        String rightPart = parts[1];

        long leftV = getValue(leftPart);
        long rightV = getValue(rightPart);

        if (leftV == rightV) {
            return true;
        } else if (leftV < rightV) {
            if (changeExpression(leftPart, rightV)) { // 只调整值更小的表达式
                return true;
            } else {
                return false;
            }
        } else {
            if (changeExpression(rightPart, leftV)) {
                return true;
            } else {
                return false;
            }
        }
    }

    // 判断是否能通过添加一位数，使得表达式值变为target
    private static boolean changeExpression(String expression, long target) {
        int l = 0, r = 0; // 记录一个数的左右边界
        for (int i = 0; i <= expression.length(); i++) {
            char ch = i < expression.length() ? expression.charAt(i) : ' ';
            if (ch >= '0' && ch <= '9') { // 数字
                r++;
            } else { // 操作符
                for (int j = l; j <= r; j++) {
                    for (int digit = 0; digit <= 9; digit++) {
                        if (j == 0 && digit == 0)
                            continue;
                        String newExpr = expression.substring(0, l) + expression.substring(l, j) + (char) ('0' + digit)
                                + expression.substring(j, r) + expression.substring(r);
                        if (getValue(newExpr) == target) {
                            return true;
                        }
                    }
                }
                r++;
                l = r;
            }
        }
        return false;
    }

    // 计算表达式的值
    private static long getValue(String expression) {
        String[] adds = expression.split("\\+");
        long result = 0L;
        for (String addStr : adds) {
            String[] muls = addStr.split("\\*");
            long temp = 1L;
            for (String mulStr : muls) {
                temp *= mulStr == "" ? 1L : Long.valueOf(mulStr);
            }
            result += temp;
        }
        return result;
    }
}

考试的时候没完全ac，这是后面自己整理的一版java解法，供各位家人们参考~也可以帮我验证检查看是否存在什么问题，欢迎大家评论指导。