【LeetCode】65. Valid Number

题目描述

Validate if a given string is numeric.

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.

解题思路

可以大概结合一下编译原理中的上下文无关文法。
大概可以有如下的规则

exponent -> float + 'e' + intfloat    -> int + '.' + int | intint      -> +DIGITS | -DIGITS | DIGITSDIGITS   -> {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} 

先将string两边的空格全部strip掉，然后利用上面的规则进行判断。
上面的规则不能覆盖得很完全。还需要仔细处理一些边界情况。
总共1500多个测例，要是不能run code并且wrong answer没有任何提示的话，想AC估计得很仔细……

AC代码

class Solution {public:    bool isInt(string s, bool flag = true) {        for (int i = 0; i < s.size(); ++i) {            //detect "+1" as true while "+" as false            if (flag && i == 0 && s.size() > 1 && (s[i] == '-' || s[i] == '+'))                continue;            if (!(s[i] >= '0' && s[i] <= '9'))                return false;        }        return true;    }    bool isFloat(string s) {        int pIdx = 0;        for (; pIdx < s.size(); ++pIdx) {            if (s[pIdx] == '.')                break;        }        if (pIdx >= s.size())            return isInt(s);        string leftS = s.substr(0, pIdx);        string rightS = s.substr(pIdx + 1, s.size() - pIdx - 1);        // detect "." as false        if (leftS == "" && rightS == "")            return false;        // detect "+.1" as true          if ((leftS == "+" || leftS == "-") && rightS != "")            return isInt(rightS, false);        return isInt(leftS) && isInt(rightS, false);    }    string strip(string s) {        int startIdx = 0;        for (;startIdx < s.size() - 1; ++startIdx) {            if (s[startIdx] != ' ')                break;        }        if (startIdx >= s.size())            return "";        int endIdx = s.size() - 1;        for (; endIdx > startIdx; --endIdx) {            if (s[endIdx] != ' ')                break;        }        return s.substr(startIdx, endIdx - startIdx + 1);    }    bool isNumber(string s) {        s = strip(s);        //find e        int i = 0;        for (; i < s.size(); ++i) {            if (s[i] == 'e')                break;        }        if (i >= s.size())            return isFloat(s);        if (i == 0 || i == s.size() - 1)            return false;        string leftS = s.substr(0, i);        string rightS = s.substr(i + 1, s.size() - i - 1);        return isFloat(leftS) && isInt(rightS);    }};