LeetCode: Valid Number 解题报告

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.

SOLUTION 1:

我们设置3个FLAG:

1. num

2. exp

3. dot

有以下情况：

（1). 出现了e，则前面要有digit,不能有e. 并且后面要有digit.

(2). 出现了.  那么是一个小数，那么前面不可以有.和e

(3). 出现了+, - 那么它必须是第一个，或者前一个是e，比如" 005047e+6"

实际的代码实现如下，相当的简洁。

感谢答案的提供者：http://www.ninechapter.com/solutions/valid-number/

大神哇哇哇！

 public class Solution {
     public boolean isNumber(String s) {
         if (s == null) {
             return false;
         }

         // cut the leading spaces and tail spaces.
         String sCut = s.trim();

         /*
         Some examples:
         "0" => true
         " 0.1 " => true
         "abc" => false
         "1 a" => false
         "2e10" => true
         */

         int len = sCut.length();

         boolean num = false;
         boolean exp = false;
         boolean dot = false;

         for (int i = 0; i < len; i++) {
             char c = sCut.charAt(i);
             if (c == 'e') {
                 if (!num || exp) {
                     return false;
                 }
                 exp = true;
                 num = false; // Should be: 2e2 , so there should be number follow "e"
             } else if (c <= '9' && c >= '0') {
                 num = true;
             } else if (c == '.') {
                 if (exp || dot) { // can't be: e0.2 can't be: ..
                     return false;
                 }
                 dot = true;
             } else if (c == '+' || c == '-') {
                 if (i != 0 && sCut.charAt(i - 1) != 'e') { // filter : " 005047e+6", this is true.
                     return false;
                 }
             } else {
                 // invalid character.
                 return false;
             }
         }

         return num;
     }
 }

请至主页君的GitHUB: https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/string/IsNumber.java

用正则其实也可以哦：

推荐一些较好的教程：

http://developer.51cto.com/art/200912/166310.htm

http://luolei.org/2013/09/regula-expression-simple-tutorial/

http://net.tutsplus.com/tutorials/php/regular-expressions-for-dummies-screencast-series/

http://deerchao.net/tutorials/regex/regex.htm

主页君就不写了，因为觉得正则实在是写不出来在面试时，真的太复杂了。

给个大神写好的正则的解答：

http://blog.csdn.net/fightforyourdream/article/details/12900751?reload