By Zhang, Cheng

更新日期:

文章目录
  1. 1. Valid Number

Valid Number

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
bool isNumber(string s) {
	int startIndex = 0;

	//先忽视掉空格
	for (int i = 0; i < s.size(); i++)
	{
		if (s[i] != ' ')
		{
			startIndex = i;
			break;
		}
	}

	//消掉之前的正负号
	if (s[startIndex] == '+' || s[startIndex] == '-')
		startIndex++;

	//从忽视掉空格后的字符开始判断
	bool result = false;
	bool isSpaceFound = false; //在表达式中是否发现空格
	bool isEFound = false; //在表达式中是否发现e
	bool symbolBehindEFound = false; //e之后的符号已经被定位
	bool isNumFound = false; //是否发现数字
	bool isDotFound = false;
	bool isNumBehindEFound = false; //e后的数字是否存在

	for (int i = startIndex; i < s.size(); i++)
	{
		if (s[i] == ' ')
			isSpaceFound = true;
		else if ((s[i] >= '0' && s[i] <= '9') || s[i] == 'e' || s[i] == '+' || s[i] == '-' || s[i] == '.')
		{
			if (isSpaceFound)
				break;

			if (s[i] >= '0' && s[i] <= '9')
			{
				isNumFound = true;

				if (isEFound)
					isNumBehindEFound = true;
			}

			if (s[i] == '+' || s[i] == '-')
			{
				if (!isEFound)
					break;

				if (symbolBehindEFound) //多个符号
					break;

				if (s[i - 1] != 'e') //符号前应是e
					break;

				symbolBehindEFound = true;
			}

			if (s[i] == '.') 
			{
				if (isEFound) //e后不能有.
					break;

				if (isDotFound) //.只能一次
					break;

				isDotFound = true;
			}

			if (s[i] == 'e')
			{
				if (!isNumFound) //e前必须有数字
					break;

				if (isEFound) //只能有一个e
					break;

				isEFound = true;
			}
		}
		else
			break;

		if (i == s.size() - 1)
		{
			if(isNumFound)
				result = true;

			if (isEFound && !isNumBehindEFound)
				result = false;
		}
	}

	return result;
}
文章目录
  1. 1. Valid Number