65. Valid Number

题目：

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.

Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button  to reset your code definition.

链接： http://leetcode.com/problems/valid-number/

题解：

看到这种题目第一反应就是DFA了，不过怎么构建好的DFA真的很难。参考了leetcode讨论版。 Automata的知识还要好好学习学习，希望年底前还有时间。

首先对字符串进行trim，去除前后的space。之后构建DFA。输入有五种情况

• 0 - 9
• +, -
• e
• dot
• other

其中dot有种特殊情况， 就是  1.成立，但 .不成立，所以对有没有数字使用一个boolean变量来记录。 应该还可以再简化，要再研究研究。

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {
    public boolean isNumber(String s) {
        if(s == null || s.length() == 0)
            return false;
        s = s.trim();
        int state = 0;
        boolean hasNum = false;

        for(int i = 0; i < s.length(); i++) {
            if(s.charAt(i) >= '0' && s.charAt(i) <= '9') {
                hasNum = true;
                if(state <= 2)
                    state = 2;
                else
                    state = (state <= 5) ? 5 : 7;
            } else if(s.charAt(i) == '+' || s.charAt(i) == '-') {
                if(state == 0 || state == 3)
                    state++;
                else
                    return false;
            } else if (s.charAt(i) == '.') {
                if(state <= 2)
                    state = 6;
                else
                    return false;
            } else if (s.charAt(i) == 'e') {
                if(state == 2 || (hasNum && state == 6) || state == 7)
                    state = 3;
                else
                    return false;
            } else
                return false;
        }

        return (state == 2 || state == 5 || (hasNum && state == 6) || state == 7);
    }
}

Test cases:

通过以下的例子我们可以看出，对dot我们需要额外判断，比如

"+.5e-5"  -  True

"+5."    - True

"5e-10.6"  - False  使用科学计数法以后不可以出现 dot

".5e10"  - True

".e10" - False

"." - False

"+5.e10" - True

二刷：

还是用state machine，画图的方法。我们详细地分解一下每个步骤。

1. 首先还是上面的图， 我们先对s进行trim操作，去除头尾的空格space
2. 设置一个变量hasNum来判断在string中是否曾经出现过数字，这个对于判断state 6的dot很关键
3. 从0开始遍历string，根据state machine写code，假设c为当前字符，我们考虑以下情况
   1. 当c为数字
   2. 当c为'+'或者'-'
   3. 当c为'.'
   4. 当c为'e'， 这时要注意从s6到s3这条， 这里的条件为 state = s6 && hasNum，这样才可以进入s3
   5. 其他返回false
4. 最后判断state是否在2, 5, 7以及 (state == 6 && hasNum)

Java:

Time Complexity - O(n)，Space Complexity - O(1)

public class Solution {
    public boolean isNumber(String s) {
        if (s == null || s.length() == 0) {
            return false;
        }
        s = s.trim();
        int state = 0;
        boolean hasNum = false;
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (c >= '0' && c <= '9') {
                hasNum = true;
                if (state <= 2) {
                    state = 2;
                } else {
                    state = (state <= 5) ? 5 : 7;
                }
            } else if (c == '+' || c == '-') {
                if (state == 0 || state == 3) {
                    state++;
                } else {
                    return false;
                }
            } else if (c == '.') {
                if (state <= 2) {
                    state = 6;
                } else {
                    return false;
                }
            } else if (c == 'e') {
                if (state == 2 || state == 7 || (state == 6 && hasNum)) {
                    state = 3;
                } else {
                    return false;
                }
            } else {
                return false;
            }
        }
        return state == 2 || state == 5 || state == 7 || (state == 6 && hasNum);
    }
}

三刷:

依然是画图使用state machine的方法。 上面的图有一个地方画错了， state 6的时候，不应该有一条自己连自己的链。需要找到一种更好的办法描述state 6的终止条件，和跳到state 3的条件。 state 6跳到state 3需要 hasNum + exp， 而终止时需要hasNum。

Java:

public class Solution {
    public boolean isNumber(String s) {
        if (s == null || s.length() == 0) return false;
        s = s.trim();
        int state = 0;
        boolean hasNum = false;
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (Character.isDigit(c)) {
                hasNum = true;
                if (state <= 2) state = 2;
                else if (state < 5) state = 5;
                else if (state == 6) state = 7;
            } else if (c == '.') {
                if (state < 3) state = 6;
                else return false;
            } else if (c == 'e') {
                if (state == 2 || (state == 6 && hasNum) || state == 7) state = 3;
                else return false;
            } else if (c == '+' || c == '-'){
                if (state == 0 || state == 3) state++;
                else return false;
            } else {
                return false;
            }
        }

        return state == 2 || state == 5 || state == 7 || (state == 6 && hasNum);
    }
}

Reference:

http://postimg.org/image/n7lsslmgz

https://leetcode.com/discuss/13691/c-my-thought-with-dfa

https://leetcode.com/discuss/55915/lol-hard-to-understand-but-fast-8ms

https://leetcode.com/discuss/9013/a-simple-solution-in-cpp

https://leetcode.com/discuss/26682/clear-java-solution-with-ifs

https://leetcode.com/discuss/23447/a-clean-design-solution-by-using-design-pattern

https://leetcode.com/discuss/70510/a-simple-solution-in-python-based-on-dfa

https://leetcode.com/discuss/47396/ac-java-solution-with-clear-explanation