Implement atoi to convert a string to an integer.
Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.
Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.
思路：题目不难，但是很多情况要考虑清楚，"", "+", " +0 123", "123ab2",以及越界的情况
1 class Solution { 2 public: 3 string getInteger(string &str) 4 { 5 int idx=0; 6 for(idx=0; idx<str.size(); ++idx) 7 { 8 if(str[idx]<‘0‘ || str[idx]>‘9‘)break; 9 } 10 return str.substr(0, idx); 11 } 12 int myAtoi(string str) { 13 if(str.size()==0)return 0; 14 int idx=0; 15 while(idx<str.size() && str[idx]==‘ ‘)++idx;//开头先去空格 16 str=str.substr(idx, str.size()-idx); 17 bool isActive=true; 18 if(str[0]>‘9‘ || str[0]<‘0‘) 19 { 20 if(str[0]!=‘+‘ && str[0]!=‘-‘)return 0; 21 if(str[0]==‘-‘)isActive=false; 22 str=str.substr(1, str.size()-1);//去正负号 23 } 24 string validStr=getInteger(str);//得到有效数字 25 if(validStr.size()==0)return 0; 26 if(validStr.size()>10)//int型，长度大于10肯定越界 27 { 28 if(isActive)return INT_MAX; 29 else return INT_MIN; 30 } 31 long n=0; 32 for(int idx=0; idx<validStr.size(); ++idx) 33 { 34 n=n*10+validStr[idx]-‘0‘; 35 } 36 if(!isActive)n=-n; 37 if(n>INT_MAX)return INT_MAX; 38 if(n<INT_MIN)return INT_MIN; 39 return n; 40 } 41 };