Valid Number

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.

思路：用有效自动机来标记每一种状态，在画自动机时要将每种状态严格区分开来，并作好记录。

1. bool isNumber(const char *s) {
2. int size = strlen(s);
3. if(size == 0) return false;
4. enum input{INVALID,SPACE,SIGN,DOT,DIGIT,EXP,NUMBER};
5. int state[][NUMBER] = {
6. -1,0,1,2,3,-1, // 0,empty { space,sign,dot,digit}
7. -1,-1,-1,2,3,-1, // 1,sign { dot,digit}
8. -1,-1,-1,-1,4,-1, //2,dot / e+/d {digit}
9. -1,8,-1,4,3,5, // 3,digit; {digit,space,dot,exp}
10. -1,8,-1,-1,4,5, // 4,dot+digit { digit,space,exp}
11. -1,-1,7,-1,6,-1, // 5,(digit+dot)e/E + S {sign,digit}
12. -1,8,-1,-1,6,-1, // 6, {digit,space}
13. -1,-1,-1,-1,6,-1,// 7. (digit + dot)e/E + d { digit}
14. -1,8,-1,-1,-1,-1 // space {space}
15. };
16. int state_number = 0;
18. while(*s!='\0') {
19. char cur = *s;
20. if(cur == ' ') {
21. state_number = state[state_number][SPACE];
22. }else if(cur == '+' || cur=='-') {
23. state_number = state[state_number][SIGN];
24. } else if(cur=='.') {
25. state_number = state[state_number][DOT];
26. } else if(cur == 'e' ||cur == 'E') {
27. state_number = state[state_number][EXP];
28. } else if(isdigit(cur)) {
29. state_number = state[state_number][DIGIT];
30. } else {
31. return false;
32. }
33. if(state_number == -1) return false;
34. s++;
35. }
36. return state_number == 6 || state_number == 8 || state_number==3 || state_number == 4;
37. }