Leetcode_65 Valid Number

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忽然想起又有一周没写博客了，今天没做其它的，就做了一道题，就写写这道题吧。题目：Validate if a given string is numeric.Some examples:"0" => true" 0.1 " => true"abc" => false"1 a" => false"2e10" => trueNote: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one.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.分析：其实就是考虑各种边界条件，考虑清楚就行了。边界条件：数字能够有前置空格和后置空格，数字中间不能有空格；对于'.'：最多仅仅可以出现1次，其前面能够没有数字，但后面一定要有数字；对于'e'：最多仅仅可以出现1次，其前后都必须有数字。但后面一定是整数，即不能出现'.'；对于'+'和'-'：'e'的前后都最多可以出现1次，且一定要在数字最前面出现；至于其它字符。只能是数字（0~9）。代码（有点笨的方法实现的，debug了好久）：bool leetcode_65_isnumber(char a){ if (a >= '0'&&a <= '9') { return true; } return false;}bool Leetcode_65::isNumber(string s){ int dianflag = 0, dian_num_begin_flag = 0, dian_num_end_flag = 0; //.的标志位 int eflag = 0, e_num_begin_flag = 0, e_num_end_flag = 0; //e的标志位 int jia_flag = 0, jian_flag = 0; //加号和减号的标志位 int shuzi_flag = 0, e_shuzi_flag = 0; //第一个标志位表示前面出现了空格再出现数字，第二个标志位表示先出现数字再出现空格 int after_e_flag = 0; int length = s.size(); for (int i = 0; i < length; ++i) { if (s[i] == '.') { if (dianflag == 1) { return false; } if (eflag == 1) { return false; } dianflag = 1; } else if (s[i] == 'e') { if (eflag == 1) { return false; } if (shuzi_flag == 0) { return false; } if (i == length - 1) { return false; } eflag = 1; } else if (s[i] == '+' || s[i] == '-') { if (dianflag == 1) { if (i >= 1 && s[i - 1] == 'e') { } else { return false; } } if (shuzi_flag == 0) { if (jia_flag == 1) { return false; } jia_flag = 1; continue; } else if (eflag == 1 && e_shuzi_flag == 0) { jia_flag = 1; continue; } else { return false; } } else if (s[i] == ' ') { if ((jia_flag == 1 || jian_flag == 1) && (shuzi_flag == 0)) { return false; } if (shuzi_flag == 1 || dianflag == 1) { while (s[i] == ' ') { ++i; if (i == length) { if (shuzi_flag == 0) { return false; } if (eflag == 1 && e_shuzi_flag == 0) { return false; } return true; } } return false; } else { continue; } } else if (leetcode_65_isnumber(s[i])) { if (dianflag == 0) { dian_num_begin_flag = 1; } else { dian_num_end_flag = 1; } if (eflag == 0) { e_num_begin_flag = 1; } else { e_num_end_flag = 1; } if (eflag == 1) { e_shuzi_flag = 1; } else { shuzi_flag = 1; } } else { return false; } } if (shuzi_flag == 0) { return false; } if (eflag == 1 && e_shuzi_flag == 0) { return false; } return true;}
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