[leetcode]Valid Number

很麻烦的题目。首先列出各种情况。
1.首尾空格；
2.+-
3.e
4.出现其他字符和空格

但还是经不住很多小细节：
"1."
"."
"+.1"
"+1."
"2e.3"
"1e"

但我觉得我的思路一开始是对的，就是先按照e分开，按照.分开，然后左右分别处理。只是后来发现左右不能用同样的方法处理。

不过这哥们的状态boolean方法自然是最好，只需扫描一遍：http://blog.unieagle.net/2012/11/06/leetcode%E9%A2%98%E7%9B%AE%EF%BC%9Avalid-number/

public class Solution {
    public boolean isNumber(String s) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if (s == null) return false;
        s = s.trim(); // removing empty spaces on two sides;
        int len = s.length();
        if (len == 0) return false;
        boolean hasE = false;
        int eIndex = -1;
        for (int i = 0; i < len; i++) {
            if (s.charAt(i) == 'e' || s.charAt(i) == 'E') {
                if (hasE) return false;
                else {
                    hasE = true;
                    eIndex = i;
                }
            }
        }
        
        if (hasE) {       
            String s1 = s.substring(0, eIndex);
            String s2 = s.substring(eIndex+1);           
            return isNumberWithoutE(s1) && isSignedNumber(s2);        
        }
        else {
            return isNumberWithoutE(s);
        }
    }
    
    private boolean isNumberWithoutE(String s) {
        int len = s.length();
        if (len == 0) return false;
        int start = 0;
        if (s.charAt(0) == '+' || s.charAt(0) == '-') start++;
        if (start == len) return false;
        s = s.substring(start);
        len = s.length();
        
        boolean hasDot = false;
        int dotIndex = -1;
        for (int i = 0; i < len; i++) {
            if (s.charAt(i) == '.') {
                if (hasDot) return false;
                else {
                    hasDot = true;
                    dotIndex = i;
                }
            }
        }
        
        if (hasDot) {
            String s1 = s.substring(0, dotIndex);
            String s2 = s.substring(dotIndex+1);
            if (s1.length() == 0 && s2.length() == 0) return false;
            if (s1.length() == 0) return isPureNumber(s2);
            if (s2.length() == 0) return isPureNumber(s1);
            return isPureNumber(s1) && isPureNumber(s2);
        }
        else {
            if (s.length() == 0) return false;
            return isPureNumber(s);
        }
    }
    
    private boolean isSignedNumber(String s) {
        int len = s.length();
        if (len == 0) return false;
        int start = 0;
        if (s.charAt(0) == '+' || s.charAt(0) == '-') start++;
        if (start == len) return false;
        s = s.substring(start);
        return isPureNumber(s);
    }
    
    private boolean isPureNumber(String s) {
        int len = s.length();
        if (len == 0) return false;;     
        
        for (int i = 0; i < len; i++) {
            char c = s.charAt(i);
            if (c >= '0' && c <= '9') continue;
            else {
                return false;
            }
        }
        return true;
    }
}