"Leetcode" atoi (Hard) ★

Although the topic is easy, but I submitted 10 times before, even hard bar.

Mostly I didn't think about it in many cases. And sometimes my rules are different from the ones in the answers.

The rules of the answer:

1. Leading spaces skip All "123" = 123

2. The sign should consider "+123" = 123 "123" = 123

3. Leading 0 of the number to skip "0000123" = "123"

4. A non-numeric value is encountered in the digital phase, and the number is truncated " -0000 123" = 0 "123a213" = 123

5. No valid number, return 0 "+-123" = 0

6. Number out of bounds, return maximum value 2147483647 or minimum-2147483648

Idea: First use strcpy to get a valid number part (first skip leading space, get sign, up to non-digit part)

Then judge the strcpy length as 0 or only one-+ +, returning 0. (No valid numbers)

There is a valid number that converts the obtained number to a string, and the strcmp determines whether it is the same. Different indicates a digital overflow.

If the input number is positive, the maximum value is returned, and the minimum value is returned instead.

#include <iostream>#include<vector>#include<algorithm>#include<queue>#include<stack>#include<string.h>using namespacestd;classSolution { Public:    intAtoiConst Char*str) {        Charscheck[ -];//Determine if overflow        Charstrcpy[ -];//a valid number part of the input string        intAns =0; inti =0; inticpy =0; //Remove leading spaces         while(Str[i] = =' ') {i++; }        if(Str[i] = ='+'|| Str[i] = ='-') {strcpy[icpy+ +] = str[i++]; }        //Remove the leading 0 that follows the sign         while(Str[i] = ='0') {i++; }         for(; Str[i]! =' /'; i++)        {            if('0'<= Str[i] && Str[i] <='9') {ans= ans *Ten+ (Str[i]-'0'); strcpy[icpy++] =Str[i]; }            Else            {                 Break; }} strcpy[icpy]=' /'; if(strlen (strcpy) = =0)        {            return 0; }        Else if(strlen (strcpy) = =1&& (strcpy[0] =='+'|| strcpy[0] =='-'))        {            return 0; }        if(strcpy[0] =='-') {ans=0-ans; sprintf (Scheck,"%d", ans); }        Else if(strcpy[0] =='+') {scheck[0] ='+'; sprintf (Scheck+1,"%d", ans); }        Else{sprintf (Scheck,"%d", ans); }        if(strcmp (Scheck, strcpy)! =0)        {            if(strcpy[0] =='-') {ans= -2147483648; }            Else{ans=2147483647; }        }        returnans; }};intMain () {solution S; Charstr[ -] ="2147483648"; intAns =s.atoi (str); return 0;}

"Leetcode" atoi (Hard) ★