Valid Number

Problem:

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

Validate if a given string is numeric.

Some examples:
"0" => true
" 0.1 " => true
"abc" => false
"1 a" => false
"2e10" => true

Solution 1:My Solution

算法思路：

1.用if语句分情况判断：

a.去掉字符串头尾的空格，比如str="  123 "，转变为str2="123"；

b.判断str2是否是空格，如果是则return false；

c.如果首字符是+，则去掉+，继续判断；

d.如果首字符是e或E,则return false；

e.判断字符串中是否存在空格或+或-，如果存在空格，则return false；

   如果存在+或1，则判断是否是e+或e-的组合；

f.判断是否存在除e或E以外的字母（如s），以及除+或-以外的字符（如*），如果存在，则return false；

g.判断e后面是否存在'.'，如果存在，则return false；

h.判断string是否以.e开头，如果是，则return false；

i. 判断是否为"+."或"-."或"."，如果是，则return false.

j. 否则以上情况不满足，则return true.

使用内建函数有：substr()、find()、find_last_of()、find_first_of()、isalpha()、ispunct()等

<pre name="code" class="cpp">    bool isNumber(const char *s)    {        string str(s);                /*1:去除头尾的空格*/        size_t start = str.find_first_not_of(' ');        if(start == string::npos)//如果字符为空格            return false;        size_t end = str.find_last_not_of(' ');        string str2 = str.substr(start, end - start + 1); //        cout<<"str2 is "<<str2<<endl;        if(str2[0] == '+')            str2 = str2.substr(1, str2.length() - 1);        if(str2[0] == 'e' || str2[0] == 'E')        {//            cout<<str<<" is e"<<endl;            return false;        }                /*2:判断字符串中是否存在空格和'-'*/        size_t len = str2.length();        for(int i = 0 ; i < len; i++)        {           if(str2[i] == ' ')            {//                cout<<"Some blanks exist in this string:"<<str<<endl;                return false;            }            if(i >= 1)            {                if(str2[i] == '+' || str2[i] == '-')                {                    if(str2[i - 1] != 'e')                        return false;                 }                                                   }        }                /*3:判断字符串中是否存在e和E以外的字母或者标点符号,以及是否存在多个e、多个'.'*/        int e_num = 0;        int dot_num = 0;        bool alpha = true;        for(int i = 0; i < len; i++)        {           char ch = str2[i];           if(isalpha(ch) && ch != 'e' && ch != 'E' || ispunct(ch) && ch != '.' && ch != '-' && ch != '+')            {//               cout<<"Some alphas or punctuation exist in this string:"<<str<<endl;                alpha = false;            }           if(ch == 'e' || ch == 'E')                e_num ++;           if(ch == '.')                dot_num ++;        }        if(!alpha || e_num >= 2 || dot_num >= 2)            return false;        /*5判断e后面是否有'.',如果有，则返回false*/        size_t e_index = str2.find_last_of('e');        size_t dot_index = str2.find_last_of('.');        if(e_index != string::npos && dot_index != string::npos)        {            if(e_index < dot_index)                return false;        }                /*6判断string是否是0e，如果是，则返回false*/        if(len >= 2)        {            if(str2.substr(0, 2) == ".e")                return false;                        }        if(str2 == ".")            return false;        if(str2[len - 1] == 'e' || str2[len - 1] == 'E' || str2[len - 1] == '+' || str2[len - 1] == '-')            return false;        if(str2 == "+." || str2 == "-.")            return false;        if(str2.substr(0, 2) == "-e")            return false;                return true;    }

Solution 2:有限状态自动机（DFA）转自网友

算法思路：

It's just some states changes depend on inputs only.

There 8 state in my states in my DFA. Below is my DFA transition diagram.

There are 5 kind of inputs in my DFA:

digit : number 0-9 for

+,- : operator + or -(negative or positive)

exp: e

dot: .

other: you can return false Immediately

4 final States in my DFA transition diagram : s2, s6, s7, s8

If the state change to final state at last, return true.

s2 can accept digits only : [phone] and etc

s5 can accept exp expression: +2.4e+12 3e9 and etc

s6 can accept decimals end with dot: 1. -42. and etc (careful, what if there exist only on dot)

s7 can accept decimals: +12.23, 87., 132

It is clear how DFA works in my pictures. We just need to handle the inputs, and update the state according to DFA.

    bool isNumber(string str) {        int state=0, flag=0; // flag to judge the special case "."        while(str[0]==' ')  str.erase(0,1);//delete the  prefix whitespace         while(str[str.length()-1]==' ') str.erase(str.length()-1, 1);//delete the suffix whitespace        for(int i=0; i<str.length(); i++){            if('0'<=str[i] && str[i]<='9'){                flag=1;                if(state<=2) state=2;                else state=(state<=5)?5:7;            }            else if('+'==str[i] || '-'==str[i]){                if(state==0 || state==3) state++;                else return false;            }            else if('.'==str[i]){                if(state<=2) state=6;                else return false;            }            else if('e'==str[i]){                if(flag&&(state==2 || state==6 || state==7)) state=3;                else return false;            }            else return false;        }        return (state==2 || state==5 || (flag&&state==6) || state==7);    }

心得总结
1.find_first_of()函数：可以查找一个字符串中的所有是否存在，而不限于一个字符；

2. DFA可以解决类似的满足某种限制条件的表达式；