leetCode 65.Valid Number (有效数字)

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.



Update (2015-02-10):

The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button  to reset your code definition.

思路：此题是真的有些难度的，并且最重要的是我根本就不清楚一个有效数字的规则。。

。

一个连规则都不知道的人怎么能玩好游戏呢，所以果断的挂了10次8次的。最后还是没有全然答对。

临时把别人的代码拿出来放在以下。有时间再好好研究一下：

public class Solution {
    public boolean isNumber(String s) {
        // Start typing your Java solution below
        // DO NOT write main() function
       /* 
 
       "0" => true
        "     0.1     " => true
        "abc" => false
        "1 a" => false
        "+-2e10" => true
        //*/
        //check input.
        if(s==null || s.length()==0) return false;
        int sz = s.length();
        int i=0;
 
        while(i<sz && s.charAt(i)==' ') ++i;
 
        boolean space = false;
        boolean exp = false;
        boolean dot = false;
        boolean number = false;
        boolean neg = false;
 
        for(; i<sz; i++) {
            char c = s.charAt(i);
            if(c==' ') {
                space = true;
            } else if(space==true) {
                return false;
            } else if( (c=='e' || c=='E') && exp==false && number==true) {
                exp = true;
                number = false;
                dot = true;
                neg = false;
            } else if( c=='.' && dot==false) {
                dot = true;
                neg = true;
               // number = false;
            } else if( c>='0' && c<='9') {
                number = true;
            } else if((c=='+' || c=='-') && neg==false && number==false ) {
                neg = true;
            } else {
                return false;
            }
        }
        return number;
    }
}