leetcode---Valid Number

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.

其实这道题目并不难   只要 考虑好  几个 特殊字符出现的位置的条件就好了  不需要  他复杂的考虑  

public class Solution {      public boolean isNumber(String s) {          s = s.trim();  //去掉两头的空格   真是醉了        if (s.length() == 0) return false;          boolean hasE = false;          boolean hasDot = false;          boolean hasNumber = false;                    for (int i = 0; i < s.length(); i++) {              // e cannot be the first character              if (i == 0 && s.charAt(i) == 'e') return false;              if (s.charAt(i) == 'e') {  //e字符不能够有两个  以及不能够在数字没出现之前出现e                // e cannot be replicated nor placed before number                  if (hasE == true || hasNumber == false) {                      return false;                  } else {                      hasE = true;                  }              }                             if (s.charAt(i) == '.') {                  // '.' cannot be replicated nor placed after 'e'  小数点 要在数字出现的后面并且不能够在E出现的后面出现                if (hasDot == true || hasE == true) {                      return false;                  } else {                      hasDot = true;                  }              }              // the sign can be placed at the beginning or after 'e'  正负号只能出现在开始以及E的后面一位            if (i != 0 && s.charAt(i - 1) != 'e' && (s.charAt(i) == '+' || s.charAt(i) == '-')) return false;                            // no other chacraters except '+', '-', '.', and 'e'              if ((s.charAt(i) > '9' || s.charAt(i) < '0') && s.charAt(i) != '+' && s.charAt(i) != '-' && s.charAt(i) != '.' && s.charAt(i) != 'e')              return false;                              // check whether numbers are included.              if (s.charAt(i) <= '9' && s.charAt(i) >= '0') {                  hasNumber = true;              }          }          // '+', '-', and 'e' cannot be the last character          if (s.charAt(s.length() - 1) == '-' || s.charAt(s.length() - 1) == '+' || s.charAt(s.length() - 1) == 'e') return false;            return hasNumber;      }  }