[leetcode]65. Valid Number 有效数值

Validate if a given string can be interpreted as a decimal number.

Some examples:
"0" => true
" 0.1 " => true
"abc" => false
"1 a" => false
"2e10" => true
" -90e3 " => true
" 1e" => false
"e3" => false
" 6e-1" => true
" 99e2.5 " => false
"53.5e93" => true
" --6 " => false
"-+3" => false
"95a54e53" => false

Note: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one. However, here is a list of characters that can be in a valid decimal number:

Numbers 0-9
Exponent - "e"
Positive/negative sign - "+"/"-"
Decimal point - "."

Of course, the context of these characters also matters in the input.

思路

There is no complex algorithem, just checking each char from input String

代码

 // {前缀空格}{浮点数}{e}{正数}{后缀空格}

 class Solution {

     public boolean isNumber(String s) {

         int len = s.length();

         int i = 0;

         int right = len - 1;

         // delete white spaces on both sides

         while (i <= right && Character.isWhitespace(s.charAt(i))) i++;

         if (i > len - 1) return false;

         while (i <= right && Character.isWhitespace(s.charAt(right))) right--;

         // check +/- sign

         if (s.charAt(i) == '+' || s.charAt(i) == '-') i++;

         boolean num = false; // is a digit

         boolean dot = false; // is a '.'

         boolean exp = false; // is a 'e'

         while (i <= right) {

             char c = s.charAt(i);

             // first char should be digit

             if (Character.isDigit(c)) {

                 num = true;

             }

             else if (c == '.') {

                 // exp and dot cannot before '.'

                 if(exp || dot) return false;

                 dot = true;

             }

             else if (c == 'e') {

                 // only one 'e'can exist

                 // 'e' should after num

                 if(exp || num == false) return false;

                 exp = true;

                 // after 'e' should exist num, so set num as default false

                 num = false;

             }

             else if (c == '+' || c == '-') {

                 if (s.charAt(i - 1) != 'e') return false;

             }

             else {

                 return false;

             }

             i++;

         }

         return num;

     }

 }

变种之简化版本的Valid Number : checking +/- numbers including decimal numbers

需要跟面试官confirm

以下哪些算valid，若以下test case中，蓝色字体是valid的

123
-123
123.123
-123.123
.123
-.123
123.
123-.

那么：

1. Decimal point 前面必须是数字

2. Decimal point 后面必须是数字

3. 整个valid number最后是以数字结尾

 public boolean isNumber(String s) {

         int len = s.length();

         int i = 0;

         int right = len - 1;

         // delete white spaces on both sides

         while (i <= right && Character.isWhitespace(s.charAt(i))) i++;

         if (i > len - 1) return false;

         while (i <= right && Character.isWhitespace(s.charAt(right))) right--;

         // check +/- sign

         if (s.charAt(i) == '+' || s.charAt(i) == '-') i++;

         boolean num = false; // is a digit

         boolean dot = false; // is a '.'

         while (i <= right) {

             char c = s.charAt(i);

             if (Character.isDigit(c)) {

                 num = true;

             }

             else if (c == '.') {

                 // (1) .. two dots  (2) no number before dot (3) - before dot

                 if( dot || num == false || s.charAt(i-1) == '-') return false;

                 dot = true;

                 // check whether there is number after decimal point

                 num = false;

             }

             else {

                 return false;

             }

             i++;

         }

         return num;

     }

巴特西

[leetcode]65. Valid Number 有效数值

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