[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;
    }