Validate if a given string can be interpreted as a decimal number.
Some examples:
"0"=>true" 0.1 "=>true"abc"=>false"1 a"=>false"2e10"=>true" -90e3 "=>true" 1e"=>false"e3"=>false" 6e-1"=>true" 99e2.5 "=>false"53.5e93"=>true" --6 "=>false"-+3"=>false"95a54e53"=>false
Note: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one. However, here is a list of characters that can be in a valid decimal number:
- Numbers 0-9
- Exponent - "e"
- Positive/negative sign - "+"/"-"
- Decimal point - "."
Of course, the context of these characters also matters in the input.
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.
有效数字。题意是给你一个input字符串，请你判断他是否为有效数字。题目没有给出非常具体的有效数字的定义，但是给了一些例子帮你理解什么叫有效什么叫无效。
思路只能是从例子下手了。注意到这么几点
- 首先input字符串里面一定需要有数字
- 之后需要判断是否有e，如果有e，则必须先看到数字才行，否则就是错的
- 如果碰到小数点，但是是在e之后或者已经出现过小数点了，就是错的
- 如果遇到加减号，看一下是不是在input的首位或者在不在e之后的那一位上，如果不满足则是错的
- 最后判断有没有遇到数字
照着这个思路，代码就不难写了。但是有的case，讨论区很多同学都不认同，我也在这里分享出来。
test(1, "123", true); test(2, " 123 ", true); test(3, "0", true); test(4, "0123", true); //Cannot agree test(5, "00", true); //Cannot agree test(6, "-10", true); test(7, "-0", true); test(8, "123.5", true); test(9, "123.000000", true); test(10, "-500.777", true); test(11, "0.0000001", true); test(12, "0.00000", true); test(13, "0.", true); //Cannot be more disagree!!! test(14, "00.5", true); //Strongly cannot agree test(15, "123e1", true); test(16, "1.23e10", true); test(17, "0.5e-10", true); test(18, "1.0e4.5", false); test(19, "0.5e04", true); test(20, "12 3", false); test(21, "1a3", false); test(22, "", false); test(23, " ", false); test(24, null, false); test(25, ".1", true); //Ok, if you say so test(26, ".", false); test(27, "2e0", true); //Really?! test(28, "+.8", true); test(29, " 005047e+6", true); //Damn = =|||
时间O(n)
空间O(1)
Java实现
1 class Solution {
2 public boolean isNumber(String s) {
3 boolean eSeen = false;
4 boolean numSeen = false;
5 boolean dotSeen = false;
6 s = s.trim();
7 for (int i = 0; i < s.length(); i++) {
8 char c = s.charAt(i);
9 if (Character.isDigit(c)) {
10 numSeen = true;
11 } else if (c == 'e') {
12 if (eSeen || !numSeen) {
13 return false;
14 }
15 eSeen = true;
16 numSeen = false;
17 } else if (c == '.') {
18 if (eSeen || dotSeen) {
19 return false;
20 }
21 dotSeen = true;
22 } else if (c == '-' || c == '+') {
23 if (i != 0 && s.charAt(i - 1) != 'e') {
24 return false;
25 }
26 } else {
27 return false;
28 }
29 }
30 return numSeen;
31 }
32 }