[LeetCode] Valid Number

这里看来的好的解答

public boolean isNumber(String s) {          if(s.trim().isEmpty()){              return false;          }          /*        re* Matches 0 or more occurrences of preceding expression.        re+ Matches 1 or more of the previous thing        re? Matches 0 or 1 occurrence of preceding expression.        a| b    Matches either a or b.        [...]   Matches any single character in brackets.        \d  Matches digits. Equivalent to [0-9].        */        String regex = "[-+]?(\\d+\\.?|\\.\\d+)\\d*(e[-+]?\\d+)?";          if(s.trim().matches(regex)){              return true;          }else{              return false;          }      }  

九章算法给了一个non regex解答

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;    }}