[LeetCode] Valid Number

Thought: 
There is a very clear solution using state machines.

Code:
public class Solution {
    public boolean isNumber(String s) {
        char[] tmp = s.toCharArray();
        int[][] trans = {
            { 0 ,0 ,0 ,0 ,0 ,0 },// false
            { 0 ,2 ,3 ,0 ,1 ,4 },// 1
            { 0 ,2 ,5 ,6 ,9 ,0 },// 2
            { 0 ,5 ,0 ,0 ,0 ,0 },// 3
            { 0 ,2 ,3 ,0 ,0 ,0 },// 4
            { 0 ,5 ,0 ,6 ,9 ,0 },// 5
            { 0 ,7 ,0 ,0 ,0 ,8 },// 6
            { 0 ,7 ,0 ,0 ,9 ,0 },// 7
            { 0 ,7 ,0 ,0 ,0 ,0 },// 8
            { 0 ,0 ,0 ,0 ,9 ,0 } // 9
        };
        int i = 0;
        int stat = 1;
        while (i < tmp.length) {
            int type = 0;
            if (tmp[i] >= '0' && tmp[i] <= '9') type = 1;
            else if (tmp[i] == '.') type = 2;
            else if (tmp[i] == 'e') type = 3;
            else if (tmp[i] == ' ') type = 4;
            else if (tmp[i] == '+' || tmp[i] == '-') type = 5;
            stat = trans[stat][type];
            i++;
        }
        return stat == 2 || stat == 5 || stat == 7 || stat == 9;
    }
}

Note:

	type	0	1	2	3	4	5
stat		others	digits	point	e	space	sign
0	FALSE
1	only space
2	digits
3	only point
4	sign
5	digits.
6	e
7	e digits
8	e sign
9	valid space