判断一个字符串是否是数字 正则表达式 自动机

我的解法

public class Solution {
    public boolean isNumeric(char[] str) {
        if(str.length==0) return false;
        int e=0;
        int point=0;
        for(int i=0;i<str.length;i++){
            if((str[i]=='+'||str[i]=='-')&&str.length>1&&(i==0||i!=0&&(str[i-1]=='e'||str[i-1]=='E'))){
                continue;
            }
            if(str[i]=='.'&&i!=0&&point==0&&e==0){
                point=1;
                continue;
            }
            if((str[i]=='e'||str[i]=='E')&&i!=0&&i!=str.length-1&&e==0){
                e=1;
                continue;
            }
            if(i==0&&str[i]=='0'&&str.length>1&&str[i+1]=='.'){
                continue;
            }
            if(str[i]>='0'&&str[i]<='9'){
                continue;
            }
            return false;
        }
        return true;
    }
}

正则表达式

public class Solution {
    public boolean isNumeric( char [] str) {
        String string = String.valueOf(str);
        return string.matches( "[\\+-]?[0-9]*(\\.[0-9]*)?([eE][\\+-]?[0-9]+)?" );
    }
}

[]可有可无


编译原理中自动机可以实现，识别如下图所示：

class Solution {
public :
    bool isNumeric( char * string)
    {
        int i = 0;
        if (string[i]== '+' || string[i]== '-' || IsNum(string[i])){
            while (string[++i]!= '\0' && IsNum(string[i]));
            if (string[i]== '.' ){
                if (IsNum(string[++i])){
                    while (string[++i]!= '\0' && IsNum(string[i]));
                    if (string[i]== 'e' ||string[i]== 'E' ){
                        i++;
                        if (string[i]== '+' || string[i]== '-' || IsNum(string[i])){
                            while (string[++i]!= '\0' && IsNum(string[i]));
                            if (string[i]== '\0' ) return true ;
                            else return false ;
                        } else return false ;
                    } else if (string[i]== '\0' ) return true ;
                    else return false ;
                } else if (string[++i]== '\0' ) return true ;
                else return false ;
            } else if (string[i]== 'e' ||string[i]== 'E' ){
                i++;
                if (string[i]== '+' || string[i]== '-' || IsNum(string[i])){
                    while (string[++i]!= '\0' && IsNum(string[i]));
                    if (string[i]== '\0' ) return true ;
                    else return false ;
                } else return false ;
            } else if (string[i]== '\0' ) return true ;
            else return false ;
        } else return false ;
    }

    bool IsNum( char ch)
    {
        if (ch< '0' ||ch> '9' ) return false ;
        else return true ;
    }
};