Valid Number

Hard String LinkedIn

LintCode: http://www.lintcode.com/en/problem/valid-number/

LeetCode: https://leetcode.com/problems/valid-number/#/description

Question

Validate if a given string is numeric.

Example
“0” => true
” 0.1 “ => true
“abc” => false
“1 a” => false
“2e10” => true

Thinking

This question is not hard. The main thing is that you need to clarify what is a numeric since the question is no clear.

• Empty string - false
• Single character “.”, “+”, “-“ - false
• Starts with “e” or ends with “e” - false
• “e” must behind dot if dot exists.
• only one “e” and one “.”
• ”+”, “-“ must be the first character if they exist.

Solution

Java

public class Solution {
    /**
     * @param s the string that represents a number
     * @return whether the string is a valid number
     */
    public boolean isNumber(String s) {
        // Write your code here
        if (s == null) {
            return false;
        }
        s = s.trim();
        // special case only one dot or empty string
        if (s.length() == 0 || s.equals(".") || s.equals("+") || s.equals("-")) {
            return false;
        }
        // special cases (start or end with 'e')
        if (s.charAt(0) == 'e' || s.charAt(s.length() - 1) == 'e') {
            return false;
        }
        boolean hasDot = false;
        boolean hasE = false;
        for (int i = 0; i < s.length(); i++) {
            char cstr = s.charAt(i);
            if (cstr == '.') {
                if (hasDot || hasE) { // e must behind dot
                    return false;
                }
                hasDot = true;
            } else if (cstr == 'e') {
                if (hasE) {
                    return false;
                }
                hasE = true;
            } else if (cstr >= '0' && cstr <= '9') {
                continue;
            } else if (cstr == '+' || cstr == '-') {
                if (i > 0) {
                    return false;
                }
            } else {
                return false;
            }
        }
        return true;
    }
}