Valid Number leetcode java

题目：

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.

题解：

正则表达式。

 本文代码引用自：http://blog.csdn.net/fightforyourdream/article/details/12900751

代码：

 1     public boolean isNumber(String s) {
 2         if(s.trim().isEmpty()){  
 3             return false;  
 4         }  
 5         String regex = "[-+]?(\\d+\\.?|\\.\\d+)\\d*(e[-+]?\\d+)?";  
 6         if(s.trim().matches(regex)){  
 7             return true;  
 8         }else{  
 9             return false;  
         }  
     }

如果按照判断的方法可以如下：

 1 public static boolean isNumber(String s) {
 2         int i = 0;
 3         while(s.charAt(i) == ' '){    // 移除前导whitespace
 4             i++;
 5             if(i >= s.length()){
 6                 return false;
 7             }
 8         }
 9         if(s.charAt(i)=='+' || s.charAt(i)=='-'){    // 忽略符号位
             i++;
         }
         int j = s.length()-1;
         while(s.charAt(j) == ' '){    // 移除后缀whitespace
             j--;
         }
         if(i <= j){
             s = s.substring(i, j+1);
         }else{
             return false;
         }
         
         int dot = -1;    // 记录点的位置
         int ee = -1;    // 记录e的位置
         for(i=0; i<s.length(); i++){
             if(dot==-1 && s.charAt(i)=='.'){
                 dot = i;
             }else if(ee==-1 && s.charAt(i)=='e'){
                 ee = i;
                 if(i+1<s.length() && (s.charAt(i+1)=='-' || s.charAt(i+1)=='+')){
                     i++;
                 }
             }else{
                 if(Character.isDigit(s.charAt(i))){
                     continue;
                 }else{
                     return false;
                 }
             }
         }
         
         //xxx.xxexx
         String startStr, midStr, lastStr;
         if(dot==-1 && ee==-1){    //xxx  
             startStr = s;    // xxx
             if(startStr.length()<1){
                 return false;
             }
         }else if(dot!=-1 && ee==-1){    //xxx.yyy  
             startStr = s.substring(0, dot);    // xxx
             midStr = s.substring(dot+1);        // yyy
             if(startStr.length()<1 && midStr.length()<1){
                 return false;
             }
         }else if(dot==-1 && ee!=-1){    // xxxeyyy
             startStr = s.substring(0, ee);    // xxx
             if(startStr.length()<1){
                 return false;
             }
             if(ee+1<s.length() && (s.charAt(ee+1)=='-' || s.charAt(ee+1)=='+')){    // xxxe-zz
                 lastStr = s.substring(ee+2);    // zz
             }else{
                 lastStr = s.substring(ee+1);
             }
             if(lastStr.length() < 1){
                 return false;
             }
         }else{        //xxx.yyezz
             if(dot>ee){        // 位置不对
                 return false;
             }
             startStr = s.substring(0, dot);    // xxx
             midStr = s.substring(dot+1, ee);    // yy
             if(startStr.length()<1 && midStr.length()<1){
                 return false;
             }
             if(ee+1<s.length() && (s.charAt(ee+1)=='-' || s.charAt(ee+1)=='+')){
                 lastStr = s.substring(ee+2);    // zz
             }else{
                 lastStr = s.substring(ee+1);
             }
             if(lastStr.length() < 1){
                 return false;
             }
         }
         return true;
     }