Valid Number
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# LeetCode #65: Valid Number (C/C++).

hard

source: https://leetcode.com/problems/valid-number/
C/C++ Solution to LeetCode problem 65. Valid Number.

## Problem

A valid number can be split up into these components (in order):

A decimal number or an integer.
(Optional) An 'e' or 'E', followed by an integer.
A decimal number can be split up into these components (in order):

(Optional) A sign character (either '+' or '-').
One of the following formats:
One or more digits, followed by a dot '.'.
One or more digits, followed by a dot '.', followed by one or more digits.
A dot '.', followed by one or more digits.
An integer can be split up into these components (in order):

(Optional) A sign character (either '+' or '-').
One or more digits.

For example, all the following are valid numbers: [“2”, “0089”, “-0.1”, “+3.14”, “4.”, “-.9”, “2e10”, “-90E3”, “3e+7”, “+6e-1”, “53.5e93”, “-123.456e789”], while the following are not valid numbers: [“abc”, “1a”, “1e”, “e3”, “99e2.5”, “–6”, “-+3”, “95a54e53”].

Given a string s, return true if s is a valid number.

Input: s = “0”
Output: true

Input: s = “e”
Output: false

Input: s = “.”
Output: false

## Constraints

• 1 <= s.length <= 20
• s consists of only English letters (both uppercase and lowercase), digits (0-9), plus '+', minus '-', or dot '.'.

## Solution

My approach to this problem is to use a FSM type of solution.

• We read the string character at a time.
• For every read, we should know what characters are not allowed at that moment. For example:
• At the start, it is not allowed to have an 'e'.
• After an 'e' is not allowed to have a dot '.'
• After a number, is not allowed to have a sign '+' nor '-'.
• If we found a not allowed character, we return false.
• Once finish, if we still need a number, return false.
• Return true.

### Solution 1:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 class Solution { public: bool isNumber(string s) { if (s.size() == 1 && (s[0] < '0' || s[0] > '9')) return false; set<char> notValid({'e'}); bool hasInt = false; bool hasDot = false; bool needsNum = false; bool hasE = false; char last; for (int i=0; i<s.size(); i++) { if (notValid.find(tolower(s[i])) != notValid.end()) return false; if (s[i] >= '0' && s[i] <= '9') { needsNum = false; notValid.insert('-'); notValid.insert('+'); if (!hasDot && !hasE) hasInt = true; if (!hasE) notValid.erase('e'); continue; } if (needsNum && last != 'e') return false; else if (s[i] == '.') { hasDot = true; if (hasInt) notValid.erase('e'); else needsNum = true; notValid.insert('.'); notValid.insert('-'); notValid.insert('+'); } else if (s[i] == '+' || s[i] == '-') { notValid.insert('+'); notValid.insert('-'); if (!hasE) notValid.erase('.'); } else if (tolower(s[i]) == 'e') { needsNum = true; hasE = true; notValid.clear(); notValid.insert('e'); notValid.insert('.'); } else return false; last = tolower(s[i]); } return !needsNum; } };