Valid Number

Change-Id: I57832fea70a05dda1efa6df055a3fdd4190a4fe5

Author: applewjg

ValidNumber.java

@@ -0,0 +1,45 @@
+/*
+ Author:     Andy, nkuwjg@gmail.com
+ Date:       Jan 25, 2015
+ Problem:    Valid Number
+ Difficulty: Hard
+ Source:     https://oj.leetcode.com/problems/valid-number/
+ Notes:
+ 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: This finite-state machine solution. Learn from fuwutu & snakeDling.
+*/
+
+public class Solution {
+    public boolean isNumber(String s) {
+        int start = 0, end = s.length() -1;
+        boolean dot = false, exp = false, digit = false;
+        while (start <= end && (s.charAt(start) == ' ')) ++start;
+        while (start <= end && (s.charAt(end) == ' ')) --end;
+        if (start <= end && (s.charAt(start) == '+' || s.charAt(start) == '-')) ++start;
+        if (start > end) return false;
+        for ( ; start <= end; ++start) {
+            if (Character.isDigit(s.charAt(start))) digit = true;
+            else if (s.charAt(start) == 'e' || s.charAt(start) == 'E') {
+                if (exp == true || digit == false || start == end) return false;
+                exp = true;
+            } else if (s.charAt(start) == '.') {
+                if (dot == true || exp == true) return false;
+                if (digit == false && start == end) return false;
+                dot = true;
+            } else if (s.charAt(start) == '+' || s.charAt(start) == '-') {
+                if (start == end) return false;
+                if (s.charAt(start-1) != 'e' && s.charAt(start-1) != 'E') return false;
+            } else return false;
+        }
+        return true;
+    }
+}