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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.
Update (2015-02-10):
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.
[Solution]
状态转移图
enum InputType { INVALID, // 0 Include: Alphas, ‘(‘, ‘&‘ ans so on SPACE, // 1 SIGN, // 2 ‘+‘,‘-‘ DIGIT, // 3 numbers DOT, // 4 ‘.‘ EXPONENT, // 5 ‘e‘ ‘E‘ }; int transTable[9][6] = { //0INVA,1SPA,2SIG,3DI,4DO,5E {-1, 0, 3, 1, 2, -1,}, //0初始无输入或者只有space的状态 {-1, 8, -1, 1, 4, 5,}, //1输入了数字之后的状态 {-1, -1, -1, 4, -1, -1,}, //2前面无数字，只输入了Dot的状态 {-1, -1, -1, 1, 2, -1,}, //3输入了符号状态 {-1, 8, -1, 4, -1, 5,}, //4前面有数字和有dot的状态 {-1, -1, 6, 7, -1, -1,}, //5‘e‘ or ‘E‘输入后的状态 {-1, -1, -1, 7, -1, -1,}, //6输入e之后输入Sign的状态 {-1, 8, -1, 7, -1, -1,}, //7输入e后输入数字的状态 {-1, 8, -1, -1, -1, -1,}, //8前面有有效数输入之后，输入space的状态 }; bool isNumber(string s) { int state = 0, i = 0; while (s[i]) { InputType input = INVALID; if (s[i] == ‘ ‘) input = SPACE; else if (s[i] == ‘+‘ || s[i] == ‘-‘) input = SIGN; else if (isdigit(s[i])) input = DIGIT; else if (s[i] == ‘.‘) input = DOT; else if (s[i] == ‘e‘ || s[i] == ‘E‘) input = EXPONENT; state = transTable[state][input]; if (state == -1) return false; i++; } return (state == 1 || state == 4 || state == 7 || state == 8); }
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原文地址：http://www.cnblogs.com/ym65536/p/4320612.html
