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I am trying to work with the sim900, what I am trying to do, is: 1- read the serial port, 2- input everything into a string, 3- search a parameter in that string, 4- clean the string.
The code is really simple, but i cant understand what i am doing wrong.
If anyone make something similar, or knows how to do it I will be graceful.
Thank you very much
Jose Luis
String leido = " ";
void setup(){ // the Serial1 baud rate
Serial.begin(9600);
Serial1.begin(9600);
}
String leido = " ";
void setup(){
// the Serial1 baud rate
Serial.begin(9600);
Serial1.begin(9600);
}
void loop()
{
//if (Serial1.available()) { Serial.write(Serial1.read()); } // Sim900
if (Serial.available()) { Serial1.write(Serial.read()); } // pc
leido = LeerSerial();
Serial.println(leido);
if (find_text("READY",leido)==1){leido = " ";}
}
String LeerSerial(){
char character;
while(Serial1.available()) {
character = Serial1.read();
leido.concat(character);
delay (10); }
if (leido != "") { Serial1.println(leido);return leido; }
}
int find_text(String needle, String haystack) {
int foundpos = -1;
for (int i = 0; (i < haystack.length() - needle.length()); i++) {
if (haystack.substring(i,needle.length()+i) == needle) {
foundpos = 1;
}
}
return foundpos;
}
You shouldn't be using == to compare strings in C/C++, since that compares pointers. A better option is strcmp or even better strncmp, check this reference.
Back to your code, try something like this:
if (strncmp(haystack.substring(i,needle.length()+i), needle, needle.length()) == 0) {
foundpos = 1;
}
Can you get away by simply using String's indexOf() ?:
String leido = " ";
void setup() {
// the Serial1 baud rate
Serial.begin(9600);
Serial1.begin(9600);
}
void loop()
{
//if (Serial1.available()) { Serial.write(Serial1.read()); } // Sim900
if (Serial.available()) {
Serial1.write(Serial.read()); // pc
}
leido = LeerSerial();
Serial.println(leido);
if (leido.indexOf("READY") == 1) {
leido = " ";
}
}
String LeerSerial() {
char character;
while (Serial1.available()) {
character = Serial1.read();
leido.concat(character);
delay (10);
}
if (leido != "") {
Serial1.println(leido);
return leido;
}
}
Note that this assumes "READY" is always at index 1.
Maybe it's worth checking if the indexOf("READY") is greater than -1 (present in the string) ?
Given two strings s and t, determine if they are isomorphic.
Two strings are isomorphic if the characters in s can be replaced to get t.
All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character but a character may map to itself.
For example,
Given "egg", "add", return true.
Given "foo", "bar", return false.
Given "paper", "title", return true.
Note:
You may assume both s and t have the same length.
I have this solution but it is taking too much time.
Any good solution will be appreciated
public boolean isIsomorphic(String s, String t) {
String resString1="",resString2="";
HashMap<Character,Integer> hashmapS = new HashMap();
HashMap<Character,Integer> hashmapT = new HashMap();
boolean flag = false;
for(int i = 0;i<s.length();i++)
{
char chS = s.charAt(i);
char chT = t.charAt(i);
if(hashmapS.containsKey(chS))
{
resString1 = resString1 + hashmapS.get(chS);
}
else
{
resString1 = resString1 + i;
hashmapS.put(chS, i);
}
if(hashmapT.containsKey(chT))
{
resString2 = resString2 + hashmapT.get(chT);
}
else
{
resString2 = resString2 + i;
hashmapT.put(chT, i);
}
}
if(resString1.equals(resString2))
return true;
else
return false;
}
/* Time complexity = O(n)*/
public static boolean isIsomorphic (String s1 , String s2){
if (s1 == null || s2 == null){
throw new IllegalArgumentException();
}
if (s1.length() != s2.length()){
return false;
}
HashMap<Character, Character> map = new HashMap<>();
for (int i = 0 ; i < s1.length(); i++){
if (!map.containsKey(s1.charAt(i))){
if(map.containsValue(s2.charAt(i))){
return false;
}
else{
map.put(s1.charAt(i), s2.charAt(i));
}
}
else{
if( map.get(s1.charAt(i)) != s2.charAt(i)){
return false;
}
}
}
return true;
}
In your implementation, you will come to know of the answer only after processing both strings completely. While in many negative test cases, answer can be determined seeing the first violation itself.
For e.g. consider 1000 character long strings: "aa.." and "ba....". An elegant solution would have to return seeing the second character itself of two strings, as 'a' cannot map to both 'a' and 'b' here.
You may find this article helpful. It also points to a C++ based solution.
Important thing to note are:
Since number of possible elements will be max pow(2, sizeof(char)), it is helpful to keep your own hash with ASCII code being the key itself. It gives significant improvement over the use of generic hash tables.
In Case of C++, use of std::urordered_map is better than std::map and std::stl as the later one uses Balanced Binary Search trees only.
Here is another implementation but with less memory usage.
public class IsoMorphic {
private static boolean isIsomorphic(String s, String t) {
if (s.length() != t.length()) {
return false;
}
char characters1[] = new char[26];
char characters2[] = new char[26];
char array1[] = s.toCharArray();
char array2[] = t.toCharArray();
for (int i=0; i<array1.length; i++) {
char c1 = array1[i];
char c2 = array2[i];
char character1 = characters1[c1-'a'];
char character2 = characters2[c2-'a'];
if (character1 == '\0' && character2 == '\0') {
characters1[c1-'a'] = array2[i];
characters2[c2-'a'] = array1[i];
continue;
}
if (character1 == array2[i] && character2 == array1[i]) {
continue;
}
return false;
}
return true;
}
public static void main(String[] args) {
System.out.println(isIsomorphic("foo", "bar")); // false
System.out.println(isIsomorphic("bar", "foo")); // false
System.out.println(isIsomorphic("paper", "title")); // true
System.out.println(isIsomorphic("title", "paper")); // true
System.out.println(isIsomorphic("apple", "orange")); // false
System.out.println(isIsomorphic("aa", "ab")); // false
System.out.println(isIsomorphic("ab", "aa")); // false
}
}
http://www.programcreek.com/2014/05/leetcode-isomorphic-strings-java/
You should be figuring out the algorithm by yourself though.
Two words are called isomorphic if the letters in single word can be remapped to get the second word. Remapping a letter means supplanting all events of it with another letter while the requesting of the letters stays unaltered. No two letters may guide to the same letter, yet a letter may guide to itself.
public bool isomorphic(string str1, string str2)
{
if (str1.Length != str2.Length)
{
return false;
}
var str1Dictionary = new Dictionary<char, char>();
var str2Dictionary = new Dictionary<char, char>();
var length = str1.Length;
for (int i = 0; i < length; i++)
{
if (str1Dictionary.ContainsKey(str1[i]))
{
if (str1Dictionary[str1[i]] != str2[i])
{
return false;
}
}
else
{
str1Dictionary.Add(str1[i], str2[i]);
}
if (str2Dictionary.ContainsKey(str2[i]))
{
if (str2Dictionary[str2[i]] != str1[i])
{
return false;
}
}
else
{
str2Dictionary.Add(str2[i], str1[i]);
}
}
return true;
}
public class Isomorphic {
/**
* #param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
System.out.println(isIsomorphic("foo", "bar"));
System.out.println(isIsomorphic("bar", "foo"));
System.out.println(isIsomorphic("foo", "bar"));
System.out.println(isIsomorphic("bar", "foo"));
System.out.println(isIsomorphic("turtle", "tletur"));
System.out.println(isIsomorphic("tletur", "turtle"));
System.out.println(isIsomorphic("turtle", "tletur"));
System.out.println(isIsomorphic("tletur", "turtle"));
}
public static boolean isIsomorphic(String s1,String s2) {
if(s1.length()!=s2.length()) {
return false;
}
if(s1.length()==1) {
return true;
}
int c1;
int c2;
for(int i=0;i<s1.length()-1;i++) {
c1=s1.charAt(i);
c2=s1.charAt(i+1);
if(c1==c2) {
c1=s2.charAt(i);
c2=s2.charAt(i+1);
if(c1==c2) {
continue;
}else {
return false;
}
}else if(c1!=c2) {
c1=s2.charAt(i);
c2=s2.charAt(i+1);
if(c1!=c2) {
continue;
}else {
return false;
}
}
}
return true;
}
}
Comments are welcome !!
public bool IsIsomorphic(string s, string t)
{
if (s == null || s.Length <= 1) return true;
Dictionary<char, char> map = new Dictionary<char, char>();
for (int i = 0; i < s.Length; i++)
{
char a = s[i];
char b = t[i];
if (map.ContainsKey(a))
{
if (map[a]==b)
continue;
else
return false;
}
else
{
if (!map.ContainsValue(b))
map.Add(a, b);
else return false;
}
}
return true;
}
Here is my implementation...
private static boolean isIsmorphic(String string1, String string2) {
if(string1==null) return false;
if(string2==null) return false;
if(string1.length()!=string2.length())return false;
HashMap<Character,Character> map=new HashMap<>();
for(int i=0;i<string1.length();i++){
char c1=string1.charAt(i);
char c2=string2.charAt(i);
if(map.get(c1)!=null && !map.get(c1).equals(c2)){
return false;
}
map.put(c1, c2);
}
return true;
}
public class Solution {
public boolean isIsomorphic(String s, String t) {
int[] m = new int[512];
for (int i = 0; i < s.length(); i++) {
if (m[s.charAt(i)] != m[t.charAt(i)+256]) return false;
m[s.charAt(i)] = m[t.charAt(i)+256] = i+1;
}
return true;
}
}
I didn't find an answer without using Maps here, so posting my implementation which don't use additional memory.
Actually using HashMap to check if words are isomorphic is very slow on short words. On my computer using the implementation is faster up to 20 symbols in test words.
static boolean isIsomorphic(String s1, String s2) {
if (s1 == null || s2 == null) return false;
final int n = s1.length();
if (n != s2.length()) return false;
for (int i = 0; i < n; i++) {
final char c1 = s1.charAt(i);
final char c2 = s2.charAt(i);
for (int j = i + 1; j < n; j++) {
if (s1.charAt(j) == c1 && s2.charAt(j) != c2) return false;
if (s2.charAt(j) == c2 && s1.charAt(j) != c1) return false;
}
}
return true;
}
Java implementation using HashMap and HashSet. O(N) = n, O(S) = c, where c is the size of the character set.
boolean isIsomorphic(String s, String t){
HashMap<Character, Character> map = new HashMap<>();
HashSet<Character> set = new HashSet<>();
if(s.length() != t.length())
return false;
for (int i = 0; i < s.length(); i++) {
if(map.containsKey(s.charAt(i))){
if(map.get(s.charAt(i)) != t.charAt(i))
return false;
} else if(set.contains(t.charAt(i))) {
return false;
} else {
map.put(s.charAt(i), t.charAt(i));
set.add(t.charAt(i));
}
}
return true;
}
There are many different ways on how to do it. Below I provided three different ways by using a dictionary, set, and string.translate.
Here I provided three different ways how to solve Isomorphic String solution in Python.
This is the best solution I think
public boolean areIsomorphic(String s1,String s2)
{
if(s1.length()!=s2.length())
return false;
int count1[] = new int[256];
int count2[] = new int[256];
for(int i=0;i<s1.length();i++)
{
if(count1[s1.charAt(i)]!=count2[s2.charAt(i)])
return false;
else
{
count1[s1.charAt(i)]++;
count2[s2.charAt(i)]++;
}
}
return true;
}
C# soluation:
public bool isIsomorphic(String string1, String string2)
{
if (string1 == null || string2 == null)
return false;
if (string1.Length != string2.Length)
return false;
var data = new Dictionary<char, char>();
for (int i = 0; i < string1.Length; i++)
{
if (!data.ContainsKey(string1[i]))
{
if (data.ContainsValue(string2[i]))
return false;
else
data.Add(string1[i], string2[i]);
}
else
{
if (data[string1[i]] != string2[i])
return false;
}
}
return true;
}
I'm having this little problem with some easy code, basically what I'm doing is sending information via the serial port via a program I wrote in Java. The information is getting their for basic statements (IE, can turn on lights and stuff) but I'm having errors getting it to decode strings with number values send to it.
So for example, I'm sending strings that look like this
BS//:+000/+000/+000
and the decoding method I'm using looks like this.
After adding the string via this:
if (inputString.startsWith("BS//:")) //**fixed
{
inputInfoToBaseStepper(inputString);
baseStepperRunAction(baseStepperRotCount, baseStepperRotStepSize, baseStepperTime);
}
Sends it too...
void inputInfoToBaseStepper(String baseStepper)
{
baseStepperRotCount = baseStepper.substring(6,9).toInt();
baseStepperRotStepSize = baseStepper.substring(10,13).toInt();
baseStepperTime = baseStepper.substring(15,18).toInt();
}
Which should decode and run
void baseStepperRunAction (int rotations, int StepSize, int delayTime)
{
for (int rotations; rotations >=0; rotations--)
{
baseStepper.step(StepSize);
delay(delayTime);
}
}
Problem seems to be that it doesn't decode... ideas I'm sort of lost at this stage. :/
(total past of the code, I know the information is getting there, just not compiling like it should.)
#include <Stepper.h>
//#include <HardwareSerial.h>
// int intensity = 0; // led intensity this is needed just as example for this sketch
String inputString = ""; // a string to hold incoming data (this is general code you can reuse)
boolean stringComplete = false; // whether the string is complete (this is general code you can reuse)
int stepsPerRevolution = 64; //at 5.625 degrees a step
// initialize the stepper library on pins 8 through 11:
Stepper baseStepper(stepsPerRevolution, 2,3,4,5); // protocols start with //BS:
Stepper shoulderStepper(stepsPerRevolution, 6,7,8,9); // protocols start with //SS:
Stepper armStepper(stepsPerRevolution, 10,11,12,13); // protocols start with //AS:
//--------baseStepper--------//
int baseStepperRotCount = 0; //how many rotations in the for loop is needed
int baseStepperRotStepSize = 0; // how large should the steps be...
int baseStepperTime = 0; //delay time needed between each step (delay); so the stepper can do it's work.
//--------shoulderStepper--------//
int shoulderStepperRotCount =0;
void setup() {
// initialize serial: (this is general code you can reuse)
Serial.begin(115200);
}
void loop() {
// when a newline arrives:
if (stringComplete) {
//these are test if statements, they serve no purpose after the intial boot, but must be included to test the connectivity;
if (inputString.startsWith("alpha"))
{
boolean msgRecognized = true;
pinMode(13, OUTPUT);
digitalWrite(13, HIGH);
inputString = "";
stringComplete = false;
}
else if (inputString.startsWith("beta"))
{
boolean msgRecognized = true;
pinMode(13, OUTPUT);
digitalWrite(13, LOW);
inputString = "";
stringComplete = false;
}
//---------------------///
//these statements set the engines and prepare for running of the program.
if (inputString.startsWith("//BS:")) // "BS//:+000/+000/+000"
{
inputInfoToBaseStepper(inputString);
baseStepperRunAction(baseStepperRotCount, baseStepperRotStepSize, baseStepperTime);
}
else if (inputString.startsWith("//SS:"))
{
//inputInfoToShoulderStepper();
//outputConfirmed();
}
else if (inputString.startsWith("//AS:"))
{
//inputInfoToArmStepper();
// outputConfirmed();
}
if(inputString.startsWith("alp://")) { // OK is a message I know (this is general code you can reuse)
boolean msgRecognized = true;
if(inputString.substring(6,10) == "kprs") { // KeyPressed }
msgRecognized = false; // this sketch doesn't know other messages in this case command is ko (not ok)
// Prepare reply message if caller supply a message id (this is general code you can reuse)
int idPosition = inputString.indexOf("?id=");
if(idPosition != -1) {
String id = inputString.substring(idPosition + 4);
// print the reply
Serial.print("alp://rply/");
if(msgRecognized) { // this sketch doesn't know other messages in this case command is ko (not ok)
Serial.print("ok?id=");
} else {
Serial.print("ko?id=");
}
Serial.print(id);
Serial.write(255); // End of Message
Serial.flush();
}
}
// clear the string:
inputString = "";
stringComplete = false;
}
}
}
/*
Send listen messages
int index = 0;
for (index = 0; index < digitalPinListeningNum; index++) {
if(digitalPinListening[index] == true) {
int value = digitalRead(index);
if(value != digitalPinListenedValue[index]) {
digitalPinListenedValue[index] = value;
Serial.print("alp://dred/");
Serial.print(index);
Serial.print("/");
Serial.print(value);
Serial.write(255);
Serial.flush();
}
}
}
for (index = 0; index < analogPinListeningNum; index++) {
if(analogPinListening[index] == true) {
int value = analogRead(index);
if(value != analogPinListenedValue[index]) {
analogPinListenedValue[index] = value;
Serial.print("alp://ared/");
Serial.print(index);
Serial.print("/");
Serial.print(value);
Serial.write(255); // End of Message
Serial.flush();
}
}
}
} */
//this method decodes and stores inputs
void inputInfoToBaseStepper(String baseStepper)
{
baseStepperRotCount = baseStepper.substring(6,9).toInt(); // B S / / : + 0 0 0 / + 0 0 0 / + 0 0 0
baseStepperRotStepSize = baseStepper.substring(10,13).toInt();// 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8
baseStepperTime = baseStepper.substring(15,18).toInt();
}
//this method runs the base stepper off the decoded actions.
void baseStepperRunAction (int rotations, int StepSize, int delayTime)
{
for (int rotations; rotations >=0; rotations--)
{
baseStepper.step(StepSize);
delay(delayTime);
}
}
/*
SerialEvent occurs whenever a new data comes in the
hardware serial RX. This routine is run between each
time loop() runs, so using delay inside loop can delay
response. Multiple bytes of data may be available.
This is general code you can reuse.
*/
void serialEvent() {
while (Serial.available() && !stringComplete) {
// get the new byte:
char inChar = (char)Serial.read();
// add it to the inputString:
inputString += inChar;
// if the incoming character is a newline, set a flag
// so the main loop can do something about it:
if (inChar == '\n') {
stringComplete = true;
}
}
}
I wrote a solution that involved OpenProcess, EnumProcessModules, GetModuleInformation and GetModuleBaseName, but apparently EnumProcessModules and GetModuleBaseName do not exist in Windows CE! What alternative is there?
I found a way to do this with CreateToolhelp32Snapshot, Module32First, Module32Next, Process32First and Process32Next. First you have to get a list of modules, then search through the list of modules to find the desired address.
#include <Tlhelp32.h>
struct MyModuleInfo
{
BYTE* Base;
HMODULE Handle;
DWORD Size;
enum { MaxNameLen = 36 };
TCHAR Name[MaxNameLen];
};
bool GetModuleList(vector<MyModuleInfo>& moduleList)
{
HANDLE hSnapshot = CreateToolhelp32Snapshot(TH32CS_SNAPPROCESS | TH32CS_SNAPMODULE | TH32CS_GETALLMODS, 0);
if (hSnapshot == INVALID_HANDLE_VALUE)
return false;
MODULEENTRY32 moduleInfo;
moduleInfo.dwSize = sizeof(moduleInfo);
if (Module32First(hSnapshot, &moduleInfo)) do {
MyModuleInfo myInfo;
myInfo.Handle = moduleInfo.hModule;
myInfo.Base = moduleInfo.modBaseAddr;
myInfo.Size = moduleInfo.modBaseSize;
memcpy(myInfo.Name, moduleInfo.szModule, min(sizeof(myInfo.Name), sizeof(moduleInfo.szModule)));
myInfo.Name[myInfo.MaxNameLen-1] = '\0';
moduleList.push_back(myInfo);
} while (Module32Next(hSnapshot, &moduleInfo));
// The module list obtained above only contains DLLs! To get the EXE files
// also, we must call Process32First and Process32Next in a loop.
PROCESSENTRY32 processInfo;
processInfo.dwSize = sizeof(processInfo);
if (Process32First(hSnapshot, &processInfo)) do {
MyModuleInfo myInfo;
myInfo.Handle = NULL; // No handle given
myInfo.Base = (BYTE*)processInfo.th32MemoryBase;
myInfo.Size = 0x800000; // No size provided! Allow max 8 MB
memcpy(myInfo.Name, processInfo.szExeFile, min(sizeof(myInfo.Name), sizeof(processInfo.szExeFile)));
myInfo.Name[myInfo.MaxNameLen-1] = '\0';
moduleList.push_back(myInfo);
} while(Process32Next(hSnapshot, &processInfo));
// Debug output
for (int i = 0; i < (int)moduleList.size(); i++) {
MyModuleInfo& m = moduleList[i];
TRACE(_T("%-30s: 0x%08x - 0x%08x\n"), m.Name, (DWORD)m.Base, (DWORD)m.Base + m.Size);
}
CloseToolhelp32Snapshot(hSnapshot);
return true;
}
const MyModuleInfo* GetModuleForAddress(vector<MyModuleInfo>& moduleList, void* address)
{
for (int m = 0; m < (int)moduleList.size(); m++) {
const MyModuleInfo& mInfo = moduleList[m];
if (address >= mInfo.Base && address < mInfo.Base + mInfo.Size)
return &mInfo;
}
return NULL;
}
I was recently asked this question in an interview:
"How could you parse a string of the form '12345' into its integer representation 12345 without using any library functions, and regardless of language?"
I thought of two answers, but the interviewer said there was a third. Here are my two solutions:
Solution 1: Keep a dictionary which maps '1' => 1, '2' => 2, etc. Then parse the string one character at a time, look up the character in your dictionary, and multiply by place value. Sum the results.
Solution 2: Parse the string one character at a time and subtract '0' from each character. This will give you '1' - '0' = 0x1, '2' - '0' = 0x2, etc. Again, multiply by place value and sum the results.
Can anyone think of what a third solution might be?
Thanks.
I expect this is what the interviewer was after:
number = "12345"
value = 0
for digit in number: //Read most significant digit first
value = value * 10 + valueOf(digit)
This method uses far less operations than the method you outlined.
Parse the string in oposite order, use one of the two methods for parsing the single digits, multiply the accumulator by 10 then add the digit to the accumulator.
This way you don't have to calculate the place value. By multiplying the accumulator by ten every time you get the same result.
Artelius's answer is extremely concise and language independent, but for those looking for a more detailed answer with explanation as well as a C and Java implementation can check out this page:
http://www.programminginterview.com/content/strings
Scroll down (or search) to "Practice Question: Convert an ASCII encoded string into an integer."
// java version
public static int convert(String s){
if(s == null || s.length() == 0){
throw new InvalidParameterException();
}
int ret = 0;
boolean isNegtive = false;
for(int i=0;i<s.length();i++){
char c = s.charAt(i);
if( i == 0 && (c == '-')){
isNegtive = true;
continue;
}
if(c - '0' < 0 || c - '0' > 10){
throw new InvalidParameterException();
}
int tmp = c - '0';
ret *= 10;
ret += tmp;
}
return isNegtive ? (ret - ret * 2) : ret;
}
//unit test
#Test
public void testConvert() {
int v = StringToInt.convert("123");
assertEquals(v, 123);
v = StringToInt.convert("-123");
assertEquals(v, -123);
v = StringToInt.convert("0");
assertEquals(v, 0);
}
#Test(expected=InvalidParameterException.class)
public void testInvalidParameterException() {
StringToInt.convert("e123");
}
#Rule
public ExpectedException exception = ExpectedException.none();
#Test
public void testInvalidParameterException2() {
exception.expect(InvalidParameterException.class);
StringToInt.convert("-123r");
}
Keep a dictionary which maps all strings to their integer counterparts, up to some limit? Doesn't maybe make much sense, except that this probably is faster if the upper limit is small, e.g. two or three digits.
You could always try a binary search through a massive look up table of string representations!
No-one said anything about efficiency... :-)
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int nod(long);
char * myitoa(long int n, char *s);
void main()
{
long int n;
char *s;
printf("Enter n");
scanf("%ld",&n);
s=myitoa(n,s);
puts(s);
}
int nod(long int n)
{
int m=0;
while(n>0)
{
n=n/10;
m++;
}
return m;
}
char * myitoa(long int n, char *s)
{
int d,i=0;
char cd;
s=(char*)malloc(nod(n));
while(n>0)
{
d=n%10;
cd=48+d;
s[i++]=cd;
n=n/10;
}
s[i]='\0';
strrev(s);
return s;
}
This is Complete program with all conditions positive, negative without using library
import java.util.Scanner;
public class StringToInt {
public static void main(String args[]) {
String inputString;
Scanner s = new Scanner(System.in);
inputString = s.nextLine();
if (!inputString.matches("([+-]?([0-9]*[.])?[0-9]+)")) {
System.out.println("error!!!");
} else {
Double result2 = getNumber(inputString);
System.out.println("result = " + result2);
}
}
public static Double getNumber(String number) {
Double result = 0.0;
Double beforeDecimal = 0.0;
Double afterDecimal = 0.0;
Double afterDecimalCount = 0.0;
int signBit = 1;
boolean flag = false;
int count = number.length();
if (number.charAt(0) == '-') {
signBit = -1;
flag = true;
} else if (number.charAt(0) == '+') {
flag = true;
}
for (int i = 0; i < count; i++) {
if (flag && i == 0) {
continue;
}
if (afterDecimalCount == 0.0) {
if (number.charAt(i) - '.' == 0) {
afterDecimalCount++;
} else {
beforeDecimal = beforeDecimal * 10 + (number.charAt(i) - '0');
}
} else {
afterDecimal = afterDecimal * 10 + number.charAt(i) - ('0');
afterDecimalCount = afterDecimalCount * 10;
}
}
if (afterDecimalCount != 0.0) {
afterDecimal = afterDecimal / afterDecimalCount;
result = beforeDecimal + afterDecimal;
} else {
result = beforeDecimal;
}
return result * signBit;
}
}