Leetcode 65 Valid Number DFA有限状态机

发布时间：2020-12-13 21:14:03 所属栏目：PHP教程来源：网络整理

导读：Validate if a given string is numeric. Some examples: 0 = true 0.1 = abc = false 1 a = 2e10 = true Note: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one. Update

Validate if a given string is numeric.

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

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

Update (2015-02⑴0):

The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument,please click the reload button to reset your code definition.

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,1,2,⑴,3},{⑴,4},5,4,6,⑴},7,8},8} }; int now=0; for(int i=0;i<s.size();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 ; } };

（编辑：李大同）

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