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I guess some of you may have noticed that this seemingly simple problem has the lowest acceptance rate among all problems on the LeetCode OJ. Well, some of you may also complain about the annoying large number of special cases and have no idea of how to handle them systematically. Well, this link provides a nicely systematic idea, which gives a very clean and clear code. Take a look at it and you will become happy with this problem :-)
The code is rewritten as follows.
1 class Solution { 2 public: 3 bool isNumber(string s) { 4 int i = 0, n = s.length(); 5 // Skip the leading spaces 6 while (i < n && isspace(s[i])) i++; 7 // Skip the sign 8 if (s[i] == ‘+‘ || s[i] == ‘-‘) i++; 9 // Check for the following significant parts 10 int digits = 0, dots = 0; 11 while (i < n && (isdigit(s[i]) || s[i] == ‘.‘)) { 12 if (isdigit(s[i])) digits++; 13 else dots++; 14 i++; 15 } 16 if (digits < 1 || dots > 1) return false; 17 if (i == n) return true; 18 // Check for the exponential parts 19 if (s[i] == ‘e‘) { 20 i++; 21 if (i == n) return false; 22 if (s[i] == ‘+‘ || s[i] == ‘-‘) i++; 23 digits = 0; 24 while (i < n && (isdigit(s[i]))) { 25 digits++; 26 i++; 27 } 28 if (digits < 1) return false; 29 } 30 // Skip the trailing spaces 31 while (i < n && isspace(s[i])) i++; 32 return i == n; 33 } 34 };
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原文地址：http://www.cnblogs.com/jcliBlogger/p/4638620.html
