提取字符串中的最长合法简单数学表达式，字符串长度最长的，并计算表达式的值。如果没有，则返回0 简单数学表达式只能包含以下内容 0-9数字，符号+-* 说明: 1.所有数字，计算结果都不超过long 2.如果有多个长度一样的，请返回第一个表达式的结果 3.数学表达式，必须是最长的，合法的 4.操作符不能连续出现，如+--+1是不合法的
输入描述:
字符串<
输出描述:
表达式值
示例1 输入输出示例仅供调试，后台判题数据一般不包含示例
输入
1-2abcd
输出
-1
Java版本
import java.util.*;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String line = sc.nextLine();
long res = 0;
int maxLen = 0;
int len = line.length();
for (int i = 0; i < len; i++) {
if (len - i <= maxLen) {
break;
}
for (int j = i; j < len; j++) {
String sub = line.substring(i, j + 1);
Matcher matcher = Pattern.compile("(\\d+)([*+-])(\\d+)").matcher(sub);
if (matcher.find() && j + 1 - i > maxLen) {
maxLen = j + 1 - i;
long first = Long.parseLong(matcher.group(1));
String op = matcher.group(2);
long second = Long.parseLong(matcher.group(3));
switch (op) {
case "+":
res = first + second;
break;
case "-":
res = first - second;
break;
case "*":
res = first * second;
break;
}
}
}
}
System.out.println(res);
}
}
Python版本
import re
line = input()
res = 0
max_len = 0
length = len(line)
for i in range(length):
if length - i <= max_len:
break
for j in range(i, length):
sub = line[i:j+1]
match = re.search(r'(\d+)([*+-])(\d+)', sub)
if match and j + 1 - i > max_len:
max_len = j + 1 - i
first = int(match.group(1))
op = match.group(2)
second = int(match.group(3))
if op == '+':
res = first + second
elif op == '-':
res = first - second
elif op == '*':
res = first * second
print(res)
C++版本
#include
#include
#include
using namespace std;
int main() {
string line;
getline(cin, line);
long long res = 0;
int maxLen = 0;
int len = line.length();
for (int i = 0; i < len; i++) {
if (len - i <= maxLen) {
break;
}
for (int j = i; j < len; j++) {
string sub = line.substr(i, j - i + 1);
smatch match;
regex pattern("(\\d+)([*+-])(\\d+)");
if (regex_search(sub, match, pattern) && j + 1 - i > maxLen) {
maxLen = j + 1 - i;
long long first = stoll(match[1]);
string op = match[2];
long long second = stoll(match[3]);
if (op == "+") {
res = first + second;
} else if (op == "-") {
res = first - second;
} else if (op == "*") {
res = first * second;
}
}
}
}
cout << res << endl;
return 0;
}
C语言版本
#include
#include
#include
bool isOperator(char c) {
return c == '+' || c == '-' || c == '*';
}
long long calculate(long long first, char op, long long second) {
if (op == '+') {
return first + second;
} else if (op == '-') {
return first - second;
} else if (op == '*') {
return first * second;
}
return 0;
}
int main() {
char line[100];
fgets(line, sizeof(line), stdin);
long long res = 0;
int maxLen = 0;
int len = strlen(line);
for (int i = 0; i < len; i++) {
if (len - i <= maxLen) {
break;
}
for (int j = i; j < len; j++) {
char sub[100];
strncpy(sub, &line[i], j - i + 1);
sub[j - i + 1] = '\0';
long long first, second;
char op;
int matched = sscanf(sub, "%lld%c%lld", &first, &op, &second);
if (matched == 3 && isOperator(op) && j + 1 - i > maxLen) {
maxLen = j + 1 - i;
res = calculate(first, op, second);
}
}
}
printf("%lld\n", res);
return 0;
}
Node.js版本
const readline = require('readline');
const rl = readline.createInterface({
input: process.stdin,
output: process.stdout
});
function isOperator(c) {
return c === '+' || c === '-' || c === '*';
}
function calculate(first, op, second) {
if (op === '+') {
return first + second;
} else if (op === '-') {
return first - second;
} else if (op === '*') {
return first * second;
}
return 0;
}
rl.question('Enter a line: ', (line) => {
let res = 0;
let maxLen = 0;
const len = line.length;
for (let i = 0; i < len; i++) {
if (len - i <= maxLen) {
break;
}
for (let j = i; j < len; j++) {
const sub = line.substring(i, j + 1);
const match = sub.match(/(\d+)([*+-])(\d+)/);
if (match && j + 1 - i > maxLen) {
maxLen = j + 1 - i;
const first = parseInt(match[1]);
const op = match[2];
const second = parseInt(match[3]);
res = calculate(first, op, second);
}
}
}
console.log(res);
rl.close();
});
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