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.

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判断数字的合法性

刚开始是把它作为1道细节较多的摹拟题做的，通过后去discuss看了1下，果然有优美的解答！

用有限状态机DFA解决，将每位看成1种状态转移条件，每次读取的1位，就根据转移矩阵进行状态转移，若转移到不合法的状态则返回false。

思路简单优美，不用斟酌过剩的细节问题，刷了这么多leetcode，这题真的眼前1亮！

具体的状态说明可以看这篇博客

class Solution {

public:

bool isNumber(string s) {

int mp[9][6]={

{⑴, 0, 1, 2, ⑴, 3},

{⑴, ⑴, ⑴, 2, ⑴, 3},

{⑴, ⑴, ⑴, ⑴, ⑴, 4},

{⑴, 5, ⑴, 4, 6, 3},

{⑴, 5, ⑴, ⑴, 6, 4},

{⑴, 5, ⑴, ⑴, ⑴, ⑴},

{⑴, ⑴, 7, ⑴, ⑴, 8},

{⑴, ⑴, ⑴, ⑴, ⑴, 8},

{⑴, 5, ⑴, ⑴, ⑴, 8}

};

int now=0;

for(int i=0;i

{

switch(s[i])

{

case '-': now=mp[now][2];break;

case '+': now=mp[now][2];break;

case ' ': now=mp[now][1];break;

case '.': now=mp[now][3];break;

case 'e': now=mp[now][4];break;

case 'E': now=mp[now][4];break;

default:

{

if(s[i]>='0' && s[i]<='9')

now=mp[now][5];

else

now=mp[now][0];

}

}

if(now==⑴) return false;

}

return now==3 || now==4 || now==5 || now==8 ;

}

};