65. Valid Number

题目：

Validate if a given string is numeric.

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

Note: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one.

解析：

本题分很多状态：

0初始无输入或者只有space的状态
1输入了数字之后的状态
2前面无数字，只输入了Dot的状态
3输入了符号状态
4前面有数字和有dot的状态
5'e' or 'E'输入后的状态
6输入e之后输入Sign的状态
7输入e后输入数字的状态
8前面有有效数输入之后，输入space的状态

共9种状态了，难设计的是6,7,8状态。

分好之后就好办了，设计出根据输入进行状态转换就OK了。

代码：

class Solutionl

{

public:

bool isNumber(const char*s)

{

enum InputType

{

INVALID,//0

SPACE,//1

SIGN,//2

DIGIT,//3

DOT,//4

EXPONENT,//5

NUM_INPUTS // 6

};

int transitionTable[][NUM_INPUTS] =

{

-1,0,3,1,2, -1,//next statesforstate0

-1,8, -1,1,4,5,//next statesforstate1

-1, -1, -1,4, -1, -1,//next statesforstate2

-1, -1, -1,1,2, -1,//next statesforstate3

-1,8, -1,4, -1,5,//next statesforstate4

-1, -1,6,7, -1, -1,//next statesforstate5

-1, -1, -1,7, -1, -1,//next statesforstate6

-1,8, -1,7, -1, -1,//next statesforstate7

-1,8, -1, -1, -1, -1,//next statesforstate8

};

intstate =0;

while (*s != '\0')

{

InputType inputType = INVALID;

if (isspace(*s))

inputType = SPACE;

elseif (*s == '+' || *s == '-')

inputType = SIGN;

elseif (isdigit(*s))

inputType = DIGIT;

elseif (*s == '.')

inputType = DOT;

elseif (*s == 'e' || *s == 'E')

inputType = EXPONENT;

state = transitionTable[state][inputType];

if (state == -1)

return false;

else

++s;

}

returnstate ==1 ||state ==4 ||state ==7 ||state ==8;

}

};