检查给定字符串是否为有效数字（整数或浮点） | SET 1（基本方法）
验证给定的字符串是否为数字。
例子：
Input : str = "11.5"
Output : true
Input : str = "abc"
Output : false
Input : str = "2e10"
Output : true
Input : 10e5.4
Output : false
以下情况需要在代码中处理。
- 忽略前导和尾随空格。
- 忽略“+”、“-”和“。”在开始时。
- 确保字符中的字符串属于 {+, -, ., e, [0-9]}
- 确保没有'.'出现在“e”之后。
- 一个点字符'.'后面应该跟一个数字。
- 字符“e”后面应该跟“+”、“-”或数字。
下面是上述步骤的实现。
C++
// C++ program to check if input number
// is a valid number
#include
#include
using namespace std;
int valid_number(string str)
{
int i = 0, j = str.length() - 1;
// Handling whitespaces
while (i < str.length() && str[i] == ' ')
i++;
while (j >= 0 && str[j] == ' ')
j--;
if (i > j)
return 0;
// if string is of length 1 and the only
// character is not a digit
if (i == j && !(str[i] >= '0' && str[i] <= '9'))
return 0;
// If the 1st char is not '+', '-', '.' or digit
if (str[i] != '.' && str[i] != '+'
&& str[i] != '-' && !(str[i] >= '0' && str[i] <= '9'))
return 0;
// To check if a '.' or 'e' is found in given
// string. We use this flag to make sure that
// either of them appear only once.
bool flagDotOrE = false;
for (i; i <= j; i++) {
// If any of the char does not belong to
// {digit, +, -, ., e}
if (str[i] != 'e' && str[i] != '.'
&& str[i] != '+' && str[i] != '-'
&& !(str[i] >= '0' && str[i] <= '9'))
return 0;
if (str[i] == '.') {
// checks if the char 'e' has already
// occurred before '.' If yes, return 0.
if (flagDotOrE == true)
return 0;
// If '.' is the last character.
if (i + 1 > str.length())
return 0;
// if '.' is not followed by a digit.
if (!(str[i + 1] >= '0' && str[i + 1] <= '9'))
return 0;
}
else if (str[i] == 'e') {
// set flagDotOrE = 1 when e is encountered.
flagDotOrE = true;
// if there is no digit before 'e'.
if (!(str[i - 1] >= '0' && str[i - 1] <= '9'))
return 0;
// If 'e' is the last Character
if (i + 1 > str.length())
return 0;
// if e is not followed either by
// '+', '-' or a digit
if (str[i + 1] != '+' && str[i + 1] != '-'
&& (str[i + 1] >= '0' && str[i] <= '9'))
return 0;
}
}
/* If the string skips all above cases, then
it is numeric*/
return 1;
}
// Driver code
int main()
{
char str[] = "0.1e10";
if (valid_number(str))
cout << "true";
else
cout << "false";
return 0;
}
// This code is contributed by rahulkumawat2107 Java
// Java program to check if input number
// is a valid number
import java.io.*;
import java.util.*;
class GFG {
public boolean isValidNumeric(String str)
{
str = str.trim(); // trims the white spaces.
if (str.length() == 0)
return false;
// if string is of length 1 and the only
// character is not a digit
if (str.length() == 1 && !Character.isDigit(str.charAt(0)))
return false;
// If the 1st char is not '+', '-', '.' or digit
if (str.charAt(0) != '+' && str.charAt(0) != '-'
&& !Character.isDigit(str.charAt(0))
&& str.charAt(0) != '.')
return false;
// To check if a '.' or 'e' is found in given
// string. We use this flag to make sure that
// either of them appear only once.
boolean flagDotOrE = false;
for (int i = 1; i < str.length(); i++) {
// If any of the char does not belong to
// {digit, +, -, ., e}
if (!Character.isDigit(str.charAt(i))
&& str.charAt(i) != 'e' && str.charAt(i) != '.'
&& str.charAt(i) != '+' && str.charAt(i) != '-')
return false;
if (str.charAt(i) == '.') {
// checks if the char 'e' has already
// occurred before '.' If yes, return 0.
if (flagDotOrE == true)
return false;
// If '.' is the last character.
if (i + 1 >= str.length())
return false;
// if '.' is not followed by a digit.
if (!Character.isDigit(str.charAt(i + 1)))
return false;
}
else if (str.charAt(i) == 'e') {
// set flagDotOrE = 1 when e is encountered.
flagDotOrE = true;
// if there is no digit before 'e'.
if (!Character.isDigit(str.charAt(i - 1)))
return false;
// If 'e' is the last Character
if (i + 1 >= str.length())
return false;
// if e is not followed either by
// '+', '-' or a digit
if (!Character.isDigit(str.charAt(i + 1))
&& str.charAt(i + 1) != '+'
&& str.charAt(i + 1) != '-')
return false;
}
}
/* If the string skips all above cases, then
it is numeric*/
return true;
}
/* Driver Function to test isValidNumeric function */
public static void main(String[] args)
{
String input = "0.1e10";
GFG gfg = new GFG();
System.out.println(gfg.isValidNumeric(input));
}
}
[/sourcecode]Python3
# Python3 program to check if input number
# is a valid number
def valid_number(str):
i = 0
j = len(str) - 1
# Handling whitespaces
while i= 0 and str[j] == ' ':
j -= 1
if i > j:
return False
# if string is of length 1 and the only
# character is not a digit
if (i == j and not(str[i] >= '0' and
str[i] <= '9')):
return False
# If the 1st char is not '+', '-', '.' or digit
if (str[i] != '.' and str[i] != '+' and
str[i] != '-' and not(str[i] >= '0' and
str[i] <= '9')):
return False
# To check if a '.' or 'e' is found in given
# string.We use this flag to make sure that
# either of them appear only once.
flagDotOrE = False
for i in range(j + 1):
# If any of the char does not belong to
# {digit, +, -,., e}
if (str[i] != 'e' and str[i] != '.' and
str[i] != '+' and str[i] != '-' and not
(str[i] >= '0' and str[i] <= '9')):
return False
if str[i] == '.':
# check if the char e has already
# occured before '.' If yes, return 0
if flagDotOrE:
return False
if i + 1 > len(str):
return False
if (not(str[i + 1] >= '0' and
str[i + 1] <= '9')):
return False
elif str[i] == 'e':
# set flagDotOrE = 1 when e is encountered.
flagDotOrE = True
# if there is no digit before e
if (not(str[i - 1] >= '0' and
str[i - 1] <= '9')):
return False
# if e is the last character
if i + 1 > len(str):
return False
# if e is not followed by
# '+', '-' or a digit
if (str[i + 1] != '+' and str[i + 1] != '-' and
(str[i + 1] >= '0' and str[i] <= '9')):
return False
# If the string skips all the
# above cases, it must be a numeric string
return True
# Driver Code
if __name__ == '__main__':
str = "0.1e10"
if valid_number(str):
print('true')
else:
print('false')
# This code is contributed by
# chaudhary_19 (Mayank Chaudhary) C#
// C# program to check if input number// is a valid numberusing System;class GFG {public Boolean isValidNumeric(String str){str = str.Trim(); // trims the white spaces.if (str.Length == 0)return false;// if string is of length 1 and the only// character is not a digitif (str.Length == 1 && !char.IsDigit(str[0]))return false;// If the 1st char is not ‘+’, ‘-‘, ‘.’ or digitif (str[0] != ‘+’ && str[0] != ‘-‘&& !char.IsDigit(str[0]) && str[0] != ‘.’)return false;// To check if a ‘.’ or ‘e’ is found in given// string. We use this flag to make sure that// either of them appear only once.Boolean flagDotOrE = false;for (int i = 1; i < str.Length; i++) {
// If any of the char does not belong to
// {digit, +, -, ., e}
if (!char.IsDigit(str[i]) && str[i] != 'e'
&& str[i] != '.' && str[i] != '+'
&& str[i] != '-')
return false;
if (str[i] == '.') {
// checks if the char 'e' has already
// occurred before '.' If yes, return 0.
if (flagDotOrE == true)
return false;
// If '.' is the last character.
if (i + 1 >= str.Length)return false;// if ‘.’ is not followed by a digit.if (!char.IsDigit(str[i + 1]))return false;}else if (str[i] == ‘e’) {// set flagDotOrE = 1 when e is encountered.flagDotOrE = true;// if there is no digit before ‘e’.if (!char.IsDigit(str[i – 1]))return false;// If ‘e’ is the last Characterif (i + 1 >= str.Length)return false;// if e is not followed either by// ‘+’, ‘-‘ or a digitif (!char.IsDigit(str[i + 1]) && str[i + 1] != ‘+’&& str[i + 1] != ‘-‘)return false;}}/* If the string skips all above cases,then it is numeric*/return true;}// Driver Codepublic static void Main(String[] args){String input = “0.1e10”;GFG gfg = new GFG();Console.WriteLine(gfg.isValidNumeric(input));}}// This code is contributed by Rajput-Ji输出：
true
时间复杂度： O(n)