LeetCode - Valid Number
Problem description
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.
Analysis
The solution for this is to use DFA to find different state and test them.
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public boolean isNumber(String s) {
s = s.trim();
State start = new State("");
State flag = new State("+-");
State numberBeforeE = new State("0123456789");
State point = new State(".");
State numberAfterPoint = new State("0123456789");
State e = new State("e");
State flagAfterE = new State("+-");
State numberAfterE = new State("0123456789");
State end = new State("");
start.next.add(flag);
start.next.add(numberBeforeE);
start.next.add(point); // really?
flag.next.add(numberBeforeE);
flag.next.add(point); // really?
numberBeforeE.next.add(point);
numberBeforeE.next.add(numberBeforeE);
numberBeforeE.next.add(e);
numberBeforeE.next.add(end);
point.next.add(numberAfterPoint);
point.next.add(end); // really?
point.next.add(e); // really?
numberAfterPoint.next.add(numberAfterPoint);
numberAfterPoint.next.add(e);
numberAfterPoint.next.add(end);
e.next.add(numberAfterE);
e.next.add(flagAfterE);
flagAfterE.next.add(numberAfterE);
numberAfterE.next.add(numberAfterE);
numberAfterE.next.add(end);
State s1 = start;
boolean hasNumberBeforeE = false;
boolean hasNumber = false;
for (char c : s.toCharArray()) {
boolean findNext = false;
for (State state : s1.next) {
if (state.context.indexOf(c) != -1) {
s1 = state;
findNext = true;
if (s1 == numberAfterE || s1 == numberAfterPoint || s1 == numberBeforeE) {
hasNumber = true;
}
if (s1 == numberBeforeE || s1 == numberAfterPoint) {
hasNumberBeforeE = true;
}
if (s1 == e && hasNumberBeforeE == false) {
return false;
}
break;
}
}
if (!findNext) {
return false;
}
}
return s1.next.contains(end) && hasNumber;
}
class State {
String context;
HashSet<State> next = new HashSet<>();
public State(String s1) {
context = s1;
}
}
What to improve
- there are some edge cases which makes it difficult to solve.
“.1” => true “3.” => true “.” => false “+.8” => true “46.e3” => true “.e1” => false “.2e81” => true