description:

看給定的數字是否是合法數字
Note:

Example:

answer:

class Solution { public:bool isNumber(string s) {int len = s.size();int left = 0, right = len - 1;bool eExisted = false;bool dotExisted = false;bool digitExisted = false;while (s[left] == ' ') ++left;while (s[right] == ' ') --right;if (left >= right && (s[left] < '0' || s[left] > '9')) return false;if (s[left] == '.') dotExisted = true;else if (s[left] >= '0' && s[left] <= '9') digitExisted = true;else if (s[left] != '+' && s[left] != '-') return false;for (int i = left + 1; i <= right - 1; ++i) {if (s[i] >= '0' && s[i] <= '9') digitExisted = true;else if (s[i] == 'e' || s[i] == 'E') {if (!eExisted && s[i - 1] != '+' && s[i - 1] != '-' && digitExisted) {eExisted = true;}else return false;} else if (s[i] == '+' || s[i] == '-') {if (s[i - 1] != 'e' && s[i - 1] != 'E') return false;} else if (s[i] == '.') {if (!dotExisted && !eExisted) dotExisted = true;else return false;} else return false;}if (s[right] >= '0' && s[right] <= '9') return true;else if (s[right] == '.' && !dotExisted && !eExisted && digitExisted) return true;else return false;} };

relative point get√：

hint :

情況很復雜
https://www.cnblogs.com/grandyang/p/4084408.html

轉載于:https://www.cnblogs.com/forPrometheus-jun/p/11374048.html

總結

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