LeetCode 65. Valid Number

原题链接在这里：https://leetcode.com/problems/valid-number/

题目：

Validate if a given string can be interpreted as a decimal number.

Some examples:
"0" => true
" 0.1 " => true
"abc" => false
"1 a" => false
"2e10" => true
" -90e3   " => true
" 1e" => false
"e3" => false
" 6e-1" => true
" 99e2.5 " => false
"53.5e93" => true
" --6 " => false
"-+3" => false
"95a54e53" => false

Note: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one. However, here is a list of characters that can be in a valid decimal number:

• Numbers 0-9
• Exponent - "e"
• Positive/negative sign - "+"/"-"
• Decimal point - "."

Of course, the context of these characters also matters in the input.

Update (2015-02-10):
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题解：

用多个flag来检测. 原string trim后从e拆开，前面部分可以有dot, 后面部分不可以有dot. 

Note: dot后面可能没有数字. After e, there must be digit.

Time Complexity: O(s.length()).

Space: O(1).

AC Java:

class Solution {
    public boolean isNumber(String s) {
        if(s == null || s.length() == 0){
            return false;
        }

        boolean eSeen = false;
        boolean dotSeen = false;
        boolean digitSeen = false;
        s = s.trim().toLowerCase();
        for(int i = 0; i < s.length(); i++){
            char c = s.charAt(i);
            if(c == '+' || c == '-'){
                if(i != 0 && s.charAt(i - 1) != 'e'){
                    return false;
                }
            }else if(Character.isDigit(c)){
                digitSeen = true;
            }else if(c == '.'){
                if(dotSeen || eSeen){
                    return false;
                }

                dotSeen = true;
            }else if(c == 'e'){
                if(eSeen || !digitSeen){
                    return false;
                }

                eSeen = true;
                digitSeen = false;
            }else{
                return false;
            }
        }

        return digitSeen;
    }
}