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.

一刷
题解：用deterministic finite automaton图来解决状态转移的问题。

1. 我们先对s进行trim操作，去除头尾的空格space
2. 设置一个变量hasNum来判断在string中是否曾经出现过数字，这个对于判断state 6的dot很关键
3. 从0开始遍历string，根据state machine写code，假设c为当前字符，我们考虑以下情况
   1. 当c为数字
   2. 当c为'+'或者'-'
   3. 当c为'.'
   4. 当c为'e'， 这时要注意从s6到s3这条， 这里的条件为 state = s6 && hasNum，这样才可以进入s3
   5. 其他返回false
4. 最后判断state是否在2, 5, 7以及 (state == 6 && hasNum)

State Machine

Time Complexity - O(n)，Space Complexity - O(1)

public class Solution {
    public boolean isNumber(String s) {
        if(s == null || s.length() == 0) return false;
        s = s.trim();
        int state = 0;
        boolean hasNum = false;
        for(int i=0; i<s.length(); i++){
            char c = s.charAt(i);
            if(c>='0' && c<='9'){//digit
                hasNum = true;
                if(state<=2) state=2;
                else state = (state<=5)? 5:7;
            }
            
            else if(s.charAt(i) == '+' || s.charAt(i)=='-'){
                if(state == 0 || state == 3) state++;
                else return false;
            }
            
            else if(s.charAt(i) == '.'){
                if(state<=2) state = 6;
                else return false;
            }
            
            else if(s.charAt(i) == 'e'){
                if(state == 2 || (hasNum && state == 6) | state ==7)
                    state = 3;
                else return false;
            }
            
            else return false;
        }
        
        return (state == 2 || state ==5 || (hasNum && state ==6) || state == 7);
    }
}