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！
1 public class Solution { 2 public boolean isNumber(String s) { 3 if (s == null) { 4 return false; 5 } 6 7 // cut the leading spaces and tail spaces. 8 String sCut = s.trim(); 9 10 /* 11 Some examples: 12 "0" => true 13 " 0.1 " => true 14 "abc" => false 15 "1 a" => false 16 "2e10" => true 17 */ 18 19 int len = sCut.length(); 20 21 boolean num = false; 22 boolean exp = false; 23 boolean dot = false; 24 25 for (int i = 0; i < len; i++) { 26 char c = sCut.charAt(i); 27 if (c == 'e') { 28 if (!num || exp) { 29 return false; 30 } 31 exp = true; 32 num = false; // Should be: 2e2 , so there should be number follow "e" 33 } else if (c <= '9' && c >= '0') { 34 num = true; 35 } else if (c == '.') { 36 if (exp || dot) { // can't be: e0.2 can't be: .. 37 return false; 38 } 39 dot = true; 40 } else if (c == '+' || c == '-') { 41 if (i != 0 && sCut.charAt(i - 1) != 'e') { // filter : " 005047e+6", this is true. 42 return false; 43 } 44 } else { 45 // invalid character. 46 return false; 47 } 48 } 49 50 return num; 51 } 52 }
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