Leetcode | | 65, Valid number

Problem:

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.

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 on the Reload button to reset your Code definition.

Hide TagsMath StringTest instructions: Determines whether a string can effectively represent a number, the representation of a number is: positive negative, Integer, fractional, scientific notation when submitted, Leetcode judgment '. 1 ' and ' 1. ' is also a valid number!! It's a little funny, it's an assessment of the system, and no one in reality thinks it works. It also caused my submission to fail.

Thinking:

(1) Details of the problem, note the ' E ' and ' e ' case

(2) Space Judgment function isspace (char*) can be called, reduce the workload

Code

By version: Refer to http://www.cnblogs.com/remlostime/archive/2012/11/18/2775938.html, note parameter char *s, not string, previous version

My own version: failed, Reason (1)leetcode judgment '. 1 ' and ' 1. ' Also a valid number (2) case ' E ' e ' does not pay attention to the distinction (3) ...

Class Solution {Public:bool Isnumber (string s) {int n=s.size ();        if (n==0) return false;            if (n==1) {if (s.at (0) >= ' 0 ' && s.at (0) <= ' 9 ') return true;        else return false;            } if (s.at (0) = = ") {string::iterator start=s.begin ();            String::iterator End=start;            while (*end== ') end++;            S.erase (Start,end);        return Isnumber (s);            } if (s.at (n-1) = = ") {string::iterator end=s.end ();            String::iterator start=end-1;            while (*start== ') start--;            S.erase (++start,end);        return Isnumber (s);            } if (s.at (0) = = ' + ' | | s.at (0) = = '-') {S.erase (S.begin (), S.begin () +1);        return Isnumber (s);        } int count1=0;        int index1=0;        int count2=0;        int index2=0; int count3=0;        for (int i=0;i<n;i++) {if ((s.at (i) >= ' 0 ' &&s.at (i) <= ' 9 ') | | s.at (i) = = '. ' | | s.at (                i) = = ' E ') {if (s.at (i) = = '. ')                    {count1++;                Index1=i;                    } else if (s.at (i) = = ' E ') {count2++;                Index2=i;            } else count3++;        } else return false;        }//for if (count3==n) return true;        if (count1>1 | | count2>1) return FALSE;            if (count2==1) {if (index2==0 | | index2==n-1) return FALSE;        else return Check2 (S,INDEX2); } if (count1==1) {if (index1==0 | |            INDEX1==N-1) return false;        else return Check1 (S,INDEX1); }}protected:bool Check1 (string s, int loc)//decimal does not contain e {for (int i=0;i<loc;i++) if (s.at (i) > ' 9 ' | | s.at (i) < ' 0 ') return FAL        Se        for (int j=loc+1;j<s.size (); j + +) if (s.at (j) > ' 9 ' | | s.at (j) < ' 0 ') return false;    return true;        } bool Check2 (string s, int loc)//Scientific count {bool flag1=true, flag2=true;        int count=0;        int index=0;            for (int i=0;i<loc;i++) {if (s.at (i) = = '. ')                {count++;                Index=i;            Continue        } if (s.at (i) < ' 0 ' | | s.at (i) > ' 9 ') return false;        } if (count>1) return false;            if (count==1) {string tmp (S,0,LOC);        if (!check1 (Tmp,index)) return false; }/* The first half of the check is finished */for (int j=loc+1;j<s.size (); j + +) {if (s.at (loc+1) = = ' + ' | | s.at (loc+1) = = '-            ') continue; if (s.at (j) < ' 0' ||        S.at (j) > ' 9 ') return false;    } return true; }};

Leetcode | | 65, Valid number