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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.
Math String
思路：分成3端，小数点前，小数点后，e后，标记为1 2 3段，1段和3段可以有正负号，1段和2段可以有任意1段为空，但不能都为空
整体上不能，看是要考虑的细节特别多
tesecase
int main() { Solution sl; #if 1 cout << sl.isNumber("0") << endl; cout << sl.isNumber("1 ") << endl; cout << sl.isNumber("1. ") << endl; cout << sl.isNumber(".1 ") << endl; cout << sl.isNumber(". ") << endl; cout << sl.isNumber(" 0.1") << endl; cout << sl.isNumber("abc") << endl; cout << sl.isNumber("1 a") << endl; cout << sl.isNumber("2.223e10") << endl; cout << sl.isNumber(" 005047e+6") << endl; cout << sl.isNumber("+++") << endl; cout << sl.isNumber("4e+") << endl; cout << sl.isNumber(".-4") << endl; #endif cout << sl.isNumber("+.8") << endl; return 0; }
最后的code
class Solution { enum ALLOW_POS_NEG {ALLOW_NONE, ALLOW_POS, ALLOW_NEG, ALLOW_ALL}; public: bool isInt(string s, enum ALLOW_POS_NEG allow = ALLOW_NONE, bool allowEmpty = false) { size_t n = s.size(); cout << "s:\t" << s << endl; int start = 0; if(allow == ALLOW_POS) { if(s[0] == ‘+‘) { start ++; } } else if(allow == ALLOW_NEG) { if(s[0] == ‘-‘) { start ++; } } else if(allow == ALLOW_ALL) { if(s[0] == ‘+‘ || s[0] == ‘-‘) { start ++; } } else //ALLOW_NONE ; if(start == n) { if(allowEmpty == true) return true; else return false; } for(int i = start; i < n; i++) { if(s[i] >= ‘0‘ && s[i] <= ‘9‘) ; else return false; } return true; } bool isNumber(string s) { size_t n = s.size(); if(n == 0) return false; int start = 0, end = n-1; // skip space while(start < n && s[start] == ‘ ‘) { start++; } // skip ‘-‘ //if(s[start] == ‘-‘ || s[start] == ‘+‘) // start ++; if(start == n) return false; // skip space while(s[end] == ‘ ‘) { end-- ; } string newStr = s.substr(start, end - start +1); size_t foundE = newStr.find(‘e‘); //cout << "newStr\t" << newStr << endl; //cout << "foundE\t" << foundE<< endl; if (foundE != string::npos)// find ‘e‘ { if(foundE == 0 || foundE == newStr.size()-1) return false; size_t foundPoint = newStr.find(‘.‘); //cout << "foundPoint\t" << foundPoint << endl; if (foundPoint != string::npos)// find ‘.‘ { if(foundPoint >= foundE) return false; if(foundPoint == 0) // ex ".1e34" return isInt(newStr.substr(foundPoint+1, foundE-foundPoint-1), ALLOW_NONE) && isInt(newStr.substr(foundE+1), ALLOW_ALL) ; if(foundPoint == foundE-1) // ex "3.e40" return isInt(newStr.substr(0, foundPoint), ALLOW_ALL) && isInt(newStr.substr(foundE+1), ALLOW_ALL) ; return isInt(newStr.substr(0, foundPoint), ALLOW_ALL, true) && isInt(newStr.substr(foundPoint+1, foundE-foundPoint-1), ALLOW_NONE) && isInt(newStr.substr(foundE+1), ALLOW_ALL) ; } else return isInt(newStr.substr(0, foundE), ALLOW_ALL) && isInt(newStr.substr(foundE+1), ALLOW_ALL); } else { size_t foundPoint = newStr.find(‘.‘); //cout << "foundPoint\t" << foundPoint << endl; if (foundPoint != string::npos)// find ‘.‘ { if(foundPoint == 0) return isInt(newStr.substr(foundPoint+1), ALLOW_NONE) ; else if(foundPoint == newStr.size()-1) return isInt(newStr.substr(0,foundPoint), ALLOW_ALL) ; else return isInt(newStr.substr(0, foundPoint), ALLOW_ALL, true) && isInt(newStr.substr(foundPoint+1), ALLOW_NONE) ; } else return isInt(newStr, ALLOW_ALL); } } };
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原文地址：http://www.cnblogs.com/diegodu/p/4323587.html
