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.
SOLUTION 1:
我们设置3个FLAG:
1. num
2. exp
3. dot
有以下情况：
（1). 出现了e，则前面要有digit,不能有e. 并且后面要有digit.
(2). 出现了. 那么是一个小数，那么前面不可以有.和e
(3). 出现了+, - 那么它必须是第一个，或者前一个是e，比如" 005047e+6"
public class Solution { public boolean isNumber(String s) { if (s == null) { return false; } // cut the leading spaces and tail spaces. String sCut = s.trim(); /* Some examples: "0" => true " 0.1 " => true "abc" => false "1 a" => false "2e10" => true */ int len = sCut.length(); boolean num = false; boolean exp = false; boolean dot = false; for (int i = 0; i < len; i++) { char c = sCut.charAt(i); if (c == 'e') { if (!num || exp) { return false; } exp = true; num = false; // Should be: 2e2 , so there should be number follow "e" } else if (c <= '9' && c >= '0') { num = true; } else if (c == '.') { if (exp || dot) { // can't be: e0.2 can't be: .. return false; } dot = true; } else if (c == '+' || c == '-') { if (i != 0 && sCut.charAt(i - 1) != 'e') { // filter : " 005047e+6", this is true. return false; } } else { // invalid character. return false; } } return num; } }