0065._Valid_Number.md

65. Valid Number

难度:Hard

刷题内容

原题连接

• https://leetcode.com/problems/valid-number/

内容描述

Validate if a given string can be interpreted as a decimal number.

Some examples:
"0" => true
" 0.1 " => true
"abc" => false
"1 a" => false
"2e10" => true
" -90e3   " => true
" 1e" => false
"e3" => false
" 6e-1" => true
" 99e2.5 " => false
"53.5e93" => true
" --6 " => false
"-+3" => false
"95a54e53" => false
Note: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one. However, here is a list of characters that can be in a valid decimal number:

Numbers 0-9
Exponent - "e"
Positive/negative sign - "+"/"-"
Decimal point - "."
Of course, the context of these characters also matters in the input.

思路 - 时间复杂度: O(N)- 空间复杂度: O(1)******

这题做了蛮久的，题目其实不难主要遍历字符串符合要求即可，但要把所有的情况都考虑进去。比如number允许前后有若干个空格，“.1”和"1.”属于合法的数字。

class Solution {
public:
    void blank(int& i,string& s)
    {
        while(i < s.length() && s[i] == ' ')
            i++;
    }
    bool firstHalf(int& i,string& s)
    {
        int num = 0;
        for(;i < s.length();++i)
            if(isdigit(s[i]))
                continue;
            else if(s[i] == '.')
            {
                if(num)
                    return 0;
                num++;
            }
            else if(s[i] == 'e' || s[i] == ' ')
                return 1;
            else
                return 0;
        return 1;
    }
    bool secondHalf(int& i,string& s)
    {
        if(i == s.length())
            return 0;
        if(s[i] =='-' || s[i] == '+')
            ++i;
        if(!isdigit(s[i]))
            return 0;
        for(;i < s.length();++i)
            if(!isdigit(s[i]))
            {
                if(s[i] == ' ')
                    return 1;
                return 0;
            }
        return 1;
    }
    bool isNumber(string s) {
        int i = 0;
        if(!s.length())
            return 0;
        blank(i,s);
        if(i == s.length())
            return 0;
        if(s[i] == '-' || s[i] == '+')
            ++i;
        int temp = i;
        if(!firstHalf(i,s))
            return 0;
        if(i == temp)
            return 0;
        if(i - 1 == temp && s[i - 1] == '.')
            return 0;
        if(i == s.length())
            return 1;
        if(s[i] == 'e')
        {
            i++;
            if(!secondHalf(i,s))
                return 0;
        }
        blank(i,s);
        return i == s.length();
    }
};