05 August 2008
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:
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.
这道题主要考察各种corner cases，包括正负号/e/点，来判断字符串是否是有效的数字。如果是在最近的面经里看到这道题可以刷一下，否则不用理会。
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class Solution {
public boolean isNumber(String s) {
s = s.trim();
boolean numberSeen = false;
boolean pointSeen = false;
boolean eSeen = false;
boolean numberAfterE = true;//表示e之后是否有number，因为最后要返回，这里初始值为true（之有数字的情况，可以过）
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) >= '0' && s.charAt(i) <= '9') {
numberSeen = true;
numberAfterE = true;//再有数字出现，再把numberAfterE赋值为true，注意第25行
} else if (s.charAt(i) == '.') {
if (eSeen || pointSeen) {//'.'可以在'e'前面，比如1.2e3；但不能在e后面，比如4e1.2就不行；另外同时出现两个'.'也不行，比如1..2
return false;
}
pointSeen = true;
} else if (s.charAt(i) == 'e') {
if (eSeen //e之前不能有另外的e
|| !numberSeen) {//e之前必须有number出现
return false;
}
eSeen = true;
numberAfterE = false;//首先有e出现，numberAfterE必为false，注意第13行
} else if (s.charAt(i) == '+' || s.charAt(i) == '-') {
if (i != 0 && s.charAt(i - 1) != 'e') {//+和-只能在第一位或者e后面出现
return false;
}
} else {
return false;
}
}
return numberSeen && numberAfterE;//必须有数字；以及如果有e的话，e后面必须跟数字
}
}