leetcode 65. Valid Number

缺乏、安全感 2022-07-29 11:39 148阅读 0赞

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.

这题虽没啥难的但要考虑的情况太多，太琐碎，所以提交了很多次，最后几乎是提交一次按错误的情况修改代码，简直要崩溃。所以建议事先收集好所有信息，在vs编辑器里定义一个数字，然后看是不是valid，这样可以避免很多错误提交。

最后写代码要耐心。

bool isNumber(char* s) {
    	//double c = 5e-5;
    	//int p = +-+-0.;
    	//double c = 3.e1;
    	//double cc = -+-+1.3e4;
    	//double v = 1.4+1e2;
    	if (s == NULL)
    		return false;
    	int len = strlen(s);
    	if (s[0] != '.'&&s[0] != ' '&&s[0] != '+'
    		&&s[0] != '-' && !(s[0] - '0' >= 0 && s[0] - '9' <= 0))
    		return false;
    	int k = 0;
    	while (k < len&&s[k] == ' ')
    		k++;
    	if (k == len)
    		return false;
    	int jj=k;
    	if (s[k] == '+' || s[k] == '-')
    	{
    		jj = k;
    		while (jj < len&&s[jj] == '+' || s[jj] == '-')
    			jj++;
    		if (jj == len)
    			return false;
    		for (int i = k; i < jj - 1; i++)
    			if (s[i] == s[i + 1])
    				return false;
    		k = jj;
    	}
    	int uu = len - 1;
    	while (uu >= 0 && s[uu] == ' ')
    		uu--;
    	for (int iii = k; iii <= uu; iii++)
    		if (s[iii] == ' ')
    			return false;
    	int dotpos = -1;
    	if (s[k] == '.')
    	{
    		int yy = k + 1;
    		while (yy < len&&s[yy] == ' ')
    			yy++;
    		if (yy == len)
    			return false;
    		dotpos = k;
    		for (int i = k + 1; i < len; i++)
    			if (s[i] == '.')
    				return false;
    	}
    	else
    	{
    		for (int i = k + 1; i < len; i++)
    			if (s[i] == '.')
    			{
    				if (dotpos >= 0)
    					return false;
    				dotpos = i;
    			}
    	}
    	if (dotpos >= 0)
    	{
    		if (dotpos == len - 1)
    		{
    			if (!(len-2>=0&&(s[len-2] - '0' >= 0 && s[len-2] - '9' <= 0)))
    				return false;
    			for (int i = jj; i < len - 1; i++)
    				if (s[i] == 'e'||s[i]=='+'||s[i]=='-')
    					return false;
    			return true;
    		}
    		for (int i = k; i < dotpos; i++)
    			if (!(s[i] - '0' >= 0 && s[i] - '9' <= 0))
    				return false;
    		int ii;
    		for (ii = dotpos + 1; ii < len; ii++)
    			if (s[ii] != ' ')
    				break;
    		if (ii == len)
    			return true;
    		k = dotpos + 1;
    	}
    	if (k == len)
    		return false;
    	if (!(s[k] - '0' >= 0 && s[k] - '9' <= 0||s[k]=='e'))
    		return false;
    	while (k < len && (s[k] - '0' >= 0 && s[k] - '9' <= 0))
    		k++;
    	if (k == len)
    		return true;
    	int j;
    	for (j = k; j < len; j++)
    		if (s[j] != ' ')
    			break;
    	if (j == len)
    		return true;
    	if (s[k] != 'e')
    		return false;
    	if (k - 1 >= 0 && s[k - 1] == '.')
    	{
    		if (!(k - 2 >= 0 && s[k - 2] - '0' >= 0 && s[k - 2] - '0' <= 9))
    			return false;
    	}
    	bool fl = false;
    	for (int i = 0; i < k; i++)
    		if (s[i] - '0' >= 0 && s[i] - '0' <= 9)
    		{
    			fl = true;
    			break;
    		}
    	if (!fl)
    		return false;
    	if (k == len - 1)
    		return false;
    	k++;
    	if (s[k] == '+' || s[k] == '-')
    		k++;
    	if (k == len)
    		return false;
    	if (!(s[k] - '0' >= 0 && s[k] - '9' <= 0))
    		return false;
    	int i;
    	for (i = k; i < len; i++)
    		if (!(s[i] - '0' >= 0 && s[i] - '9' <= 0))
    			break;
    	if (i == len)
    		return true;
    	for (; i < len; i++)
    		if (s[i] != ' ')
    			return false;
    	return true;
    }