String to INTEGER (atoi)

Problem: Convert the escape character to a number
Analysis: I think the question is not difficult, but there are many details, which are easy to think.
1. There is a space in front of the number, such as S = "123456"
2. there are unnecessary or more characters before the number, leading to a digital authentication error. The output value is 0, for example, S = "b1234", S = "++ 1233", and S = "+-1121"
3. There are unnecessary characters in the number. The number before the character is returned, for example, S = "12a12", S = "123 123"
4. The number is out of the range (-2147483648--2147483647). If the value of output-2147483648 exceeds the value of positive output 2147483647
In a knowledge point of science, if a certain number exceeds 2147483647, it will become a negative number, the opposite is the same

 

Class solution {public: int atoi (const char * Str) {long cur = 0; // initialization always forgets int num = 0, I = 0; int flag1 = 0, flag2 = 0; while (STR [I]! = '\ 0' & STR [I] = '') I ++; // discard if (STR [I] = '-') flag1 ++, I ++; else if (STR [I] = '+') flag2 ++, I ++; For (; STR [I]! = '\ 0'; I ++) {If (STR [I]> = '0' & STR [I] <= '9 ') {If (flag1 = 2) {cur = cur * 10-(STR [I]-'0'); // here is subtraction, because the cur symbol is a negative if (cur <-2147483648) Return-2147483648;} else if (flag1 = 1) cur =-str [I] + '0 ', flag1 ++; // record the negative number in the cur. else {cur = cur * 10 + (STR [I]-'0'); If (cur> 2147483647) return 2147483647 ;}} else break;} num = (INT) cur; return num ;}};