Leetcode - Valid Number

最新推荐文章于 2020-08-25 20:59:04 发布

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本文深入探讨了数值验证问题，通过构建确定性有限自动机（DFA）来解决字符串是否为数值的问题。详细介绍了如何从无效状态、整数状态、小数状态、指数状态等不同场景出发，构建状态转移表，并最终通过正则表达式实现数值验证。文章提供了直观的DFA示例和转换表格，帮助理解数值验证背后的逻辑。

https://leetcode.com/problems/valid-number/

Validate if a given string is numeric.

Some examples:

"0" => true
" 0.1 " => true
"abc" => false
"1 a" => false
"2e10" => true

Note: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one.

Some other hidden tests:

"3e3.1" =>false
" +.1 " => true
"-1." =>true

Here is a nice solution for this problem: http://blog.youkuaiyun.com/kenden23/article/details/18696083

Using Automaton!

The problem is, how to generate Deterministic Finite Automaton(DFA) ?

Assume that the string has been stripped (i.e. no spaces at head or at tail), for each state, there are 5 possible inputs: invalid, [0-9], +/-, dot(.) and e/E.

For any state, invalid input would directly lead to an invalid state (State -1). And feed each input to current state, observe whether this would lead to an invalid state (State-1) or valid one. If the output state is valid, is it a new state you have never seen before or could it be covered by previous state?

By answering 2 questions above, I formed the following DFA (only valid states are drawn):

S0: start

S1: only contains integers (this is a valid final state)

S2: only contains +/- (this is an invalid final state)

S3: contains decimal point but without any tenths number (this is an invalid final state)

S4: contains any decimal number (this is a valid final state)

S5: now we encountered exponential symbol (this is an invalid final state)

S6: ...e+/-, a temporary state for numbers like ..e-1, e+1

S7: all numbers with exponential symbol (valid final state)

From above DFA digram, we can generate the transition table. Given an input and current state, we can know the next state with this table.

Transition Table
	invalid	0..9	+/-	dot	e/E
S0	-1	1	2	3	-1
S1	-1	1	-1	4	5
S2	-1	1	-1	3	-1
S3	-1	4	-1	-1	-1
S4	-1	4	-1	-1	5
S5	-1	7	6	-1	-1
S6	-1	7	-1	-1	-1
S7	-1	7	-1	-1	-1

Code

public class Solution {
    public boolean isNumber(String s) {
        if (s == null || s.length() == 0)
            return false;
        int[][] transTable = {{-1,1,2,3,-1},{-1,1,-1,4,5},
                            {-1,1,-1,3,-1},{-1,4,-1,-1,-1},
                            {-1,4,-1,-1,5},{-1,7,6,-1,-1},
                            {-1,7,-1,-1,-1}, {-1,7,-1,-1,-1}};
        s = s.trim();
        int input = 0, state = 0;
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (c >= '0' && c <= '9')
                input = 1;
            else if (c == 'e' || c == 'E')
                input = 4;
            else if (c == '+' || c == '-')
                input = 2;
            else if (c == '.')
                input = 3;
            else input = 0;
            state = transTable[state][input];
            if (state == -1)
                return false;
        }
        return state == 1 || state == 4 || state == 7;
    }
}