leetcode 65 Valid Number ----- java

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.

这道题题目意思很简单，就是给定一个string，然后判断是否是一个数字，通过率相对很低了

主要就是判断清楚所有情况。

我在做的时候也是出了很多五错误，还是因为情况太多了，考虑的不周全。

这道题其实应该考虑一个自动机的问题。

public class Solution {
    public boolean isNumber(String s) {
        int len = s.length();     
        int flag1 = 0,flag2 = 0,flag3 = 0,flag4 = 0;
        int isnum = 0;
        for( int i = 0;i<len;i++){
            if( s.charAt(i) == ' ' && flag1 == 0 ){
                while(i<len && s.charAt(i) == ' ' )
                    i++;
                i--;
                flag1 = 1;
            }
            else if( s.charAt(i) == '.' && flag2 == 0){
                flag1 = 1;
                flag2 = 1;
            }
            else if( (s.charAt(i) == '-' || s.charAt(i) == '+') && flag3 == 0 && isnum == 0 && flag2 == 0){
                flag1 = 1;
                flag3 = 1;
            }
            else if( s.charAt(i) == 'e' && flag4 == 0 && isnum == 1) {
                flag1 = 1;
                flag4 = 1;
                flag2 = 1;
                isnum = 0;
                if( i<len-1 && (s.charAt(i+1) == '-' || s.charAt(i+1) == '+')){
                    i++;
                }
            }
            else if( s.charAt(i) >= '0' && s.charAt(i) <= '9'){
                flag1 = 1;
                isnum = 1;
            }
            else if( s.charAt(i) == ' '){
                while(i<len && s.charAt(i) == ' ' )
                    i++;
                if( i == len && isnum == 1)
                    return true;
                else
                    return false;
            }else
                return false;
            
        }
        return isnum == 1?true:false;
        
        
        
    }
}