前言

Leetcode做到Valid Number这道题， 看到 Leetcode题目分类 中说这道题难度为二级（总共五级）， 还以为这道题真的很简单。。。真是坑了大爹了。

仔细总结了下， 这道题大概有三种解法：

题目

https://oj.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.

解法一：

这种方法用正则表达式来解决， 算是最简单优雅的一种方法了。

class Solution2
{
public:
    bool isNumber(const char *s) 
    {   
<span style="white-space:pre">	</span>regex re("( *)[+-]?(\\d+|\\d+\\.\\d*|\\d*\\.\\d+)([eE][+-]?\\d+)?");

        if (regex_match(s, re))
            return true;
        else
            return false;
    }   

};

不过截止到发文章， g++ 4.8.2貌似还是没有完全支持正则表达式， 所以这段代码编译会报错。

但是如果用Java或Python的话， 应该没问题了。

另一方面， 面试的时候恐怕不会允许简简单单的用正则表达式来解决这个问题， 所以还要掌握其他方法。

解法二：

2.1 DFA

第二种方法用DFA来解决这个问题， 用到了了编译原理中的tokenizer那部分的思想。

这种方法写代码之前要先画出图来才行，我画的如下：

画的有点乱， 不过领会精神就好。 ：）

代码如下：

class Solution2_1
{
public:
public:
	bool isNumber(const char *s)
	{
		int state = 0;

		while (1)
		{
			switch (state)
			{
				case 0:
					if (*s == ' ')
						state = 0;
					else if (*s == '+' || *s == '-')
						state = 1;
					else if (isdigit(*s))
						state = 2;
					else if (*s == '.')
						state = 9;
					else 
						state = 8;
					break;

				case 1:
					if (isdigit(*s))
						state = 2;
					else if (*s == '.')
						state = 9;
					else
						state = 8;
					break;

				case 2:
					if (isdigit(*s))
						state = 2;
					else if (*s == '.')
						state = 3;
					else if (*s == '\0')
						state = 7;
					else if (*s == ' ')
						state = 10;
					else if (*s == 'e' || *s == 'E')
						state = 5;
					else
						state = 8;
					break;
					
				case 3:
					if (isdigit(*s))
						state = 4;
					else if (*s == '\0')
						state = 7;
					else if (*s == ' ')
						state = 10;
					else if (*s == 'e' || *s == 'E')
						state = 5;
					else
						state = 8;
					break;

				case 4:
					if (isdigit(*s))
						state = 4;
					else if (*s == 'e' || *s == 'E')
						state = 5;
					else if (*s == '\0')
						state = 7;
					else if (*s == ' ')
						state = 10;
					else
						state = 8;
					break;

				case 5:
					if (isdigit(*s))
						state = 6;
					else if (*s == '+' || *s == '-')
						state = 11;
					else
						state = 8;
					break;

				case 6:
					if (isdigit(*s))
						state = 6;
					else if (*s == ' ')
						state = 6;
					else if (*s == '\0')
						state = 7;
					else if (*s == ' ')
						state = 10;
					else
						state = 8;
					break;

				case 7:
					return true;

				case 8:
					return false;

				case 9:
					if (isdigit(*s))
						state = 4;
					else
						state = 8;
					break;

				case 10:
					if (*s == '\0')
						state = 7;
					else if (*s == ' ')
						state = 10;
					else
						state = 8;
					break;
				case 11:
					if (isdigit(*s))
						state = 6;
					else 
						state = 8;
					break;
			}
			s++;
		}
	}
};

这段代码可能开起来有点长， 不过只要弄明白了那张状态转换图很容易明白的。

2.2 DFA改进版

上面这段代码可能有点长， 所以用二维数组对其进行改进：

代码如下：

class Solution2_2
{
public:
	bool isNumber(const char *s)
	{
		int state = 0;
		int translate[][7] =
		{
			0, 1, 2, 8, 9, 8, 8,	//0
			8, 8, 2, 8, 9, 8, 8,	//1
			10, 8, 2, 5, 3, 7, 8,	//2
			10, 8, 4, 5, 8, 7, 8,	//3
			10, 8, 4, 5, 8, 7, 8, 	//4
			8, 11, 6, 8, 8, 8, 8,	//5
			10, 8, 6, 8, 8, 7, 8,	//6
			7, 7, 7, 7, 7, 7, 7,	//7
			8, 8, 8, 8, 8, 8, 8,	//8
			8, 8, 4, 8, 8, 8, 8,	//9
			10, 8, 8, 8, 8, 7, 8, 	//10
			8, 8, 6, 8, 8, 8, 8		//11
		};

		int type;
		while (1)
		{
			if (*s == ' ')
				type = 0;
			else if (*s == '+' || *s == '-')
				type = 1;
			else if (isdigit(*s))
				type = 2;
			else if (*s == 'e' || *s == 'E')
				type = 3;
			else if (*s == '.')
				type = 4;
			else if (*s == '\0')
				type = 5;
			else 
				type = 6;

			state = translate[state][type];

			if (state == 7)
				return true;
			else if (state == 8)
				return false;

			s++;
		}
	}
};

解法三：

用多个flag来标识现在的状态， 可读性比较差， 不推荐。

class Solution3 {
public:
    bool isNumber(const char *s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        if (s == NULL)
             return false;
             
         while(isspace(*s))
             s++;
             
         if (*s == '+' || *s == '-')
             s++;
             
         bool eAppear = false;
         bool dotAppear = false;
         bool firstPart = false;
         bool secondPart = false;
         bool spaceAppear = false;
         while(*s != '\0')
         {
             if (*s == '.')
             {
                 if (dotAppear || eAppear || spaceAppear)
                     return false;
                 else
                     dotAppear = true;
             }
             else if (*s == 'e' || *s == 'E')
             {
                 if (eAppear || !firstPart || spaceAppear)
                     return false;
                 else
                     eAppear = true;
             }
             else if (isdigit(*s))
             {
                 if (spaceAppear)
                     return false;
                     
                 if (!eAppear)
                     firstPart = true;
                 else
                     secondPart = true;
             }
             else if (*s == '+' || *s == '-')   // behind of e/E
             {
                 if (spaceAppear)
                     return false;
                     
                 if (!eAppear || !(*(s-1) == 'e' || *(s-1) == 'E'))
                     return false;
             }
             else if (isspace(*s))
                 spaceAppear = true;
             else
                 return false;
                 
             s++;            
         }
         
         if (!firstPart) {
             return false;
         }  else if (eAppear && !secondPart) {
             return false;
         } 
         return true;  
    }
};

源码：

Github上的源码

参考：

Leetcode 150题目终结贴 - Valid Number

【leetcode】Valid Number

本文章迁移自http://blog.csdn.net/timberwolf_2012/article/details/38487239