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.
解法一：
/** * 本代码由九章算法编辑提供。没有版权欢迎转发。 * - 九章算法致力于帮助更多中国人找到好的工作，教师团队均来自硅谷和国内的一线大公司在职工程师。 * - 现有的面试培训课程包括：九章算法班，系统设计班，BAT国内班 * - 更多详情请见官方网站：http://www.jiuzhang.com/ */ // Non-regex version public class Solution { public boolean isNumber(String s) { int len = s.length(); int i = 0, e = len - 1; while (i <= e && Character.isWhitespace(s.charAt(i))) i++; if (i > len - 1) return false; while (e >= i && Character.isWhitespace(s.charAt(e))) e--; // skip leading +/- if (s.charAt(i) == '+' || s.charAt(i) == '-') i++; boolean num = false; // is a digit boolean dot = false; // is a '.' boolean exp = false; // is a 'e' while (i <= e) { char c = s.charAt(i); if (Character.isDigit(c)) { num = true; } else if (c == '.') { if(exp || dot) return false; dot = true; } else if (c == 'e') { if(exp || num == false) return false; exp = true; num = false; } else if (c == '+' || c == '-') { if (s.charAt(i - 1) != 'e') return false; } else { return false; } i++; } return num; } }
解法二：
CleanCode Version: better logic
public class Solution { public boolean isNumber(String s) { int i=0; int n=s.length(); while(i<n&&Character.isWhitespace(s.charAt(i)))i++; if(i<n&&(s.charAt(i)=='+'||s.charAt(i)=='-'))i++; boolean isNumeric = false; while(i<n&&Character.isDigit(s.charAt(i))) { i++; isNumeric = true; } if(i<n&&s.charAt(i)=='.') { i++; while(i<n&&Character.isDigit(s.charAt(i))) { i++; isNumeric = true; } } if(isNumeric&&i<n&&s.charAt(i)=='e') { i++; isNumeric = false; if(i<n&&(s.charAt(i)=='+'||s.charAt(i)=='-'))i++; while(i<n&&Character.isDigit(s.charAt(i))) { i++; isNumeric = true; } } while(i<n&&Character.isWhitespace(s.charAt(i)))i++; return i==n&&isNumeric; } }
转载于:https://www.cnblogs.com/hygeia/p/5115387.html
