Leetcode 8. String to Integer (atoi) atoi函数实现 (字符串)

Input: "42"
Output: 42

Input: "   -42"
Output: -42

Input: "4193 with words"
Output: 4193

Input: "words and 987"
Output: 0

"42"
"words and 987"
"-91283472332"
"0-1"
"-000000000000001"
"  0000000000012345678"
"10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000522545459"
"-2147483647"
"-2147483648"
"2147483648"
"2147483649"
""
"7"
"   +0 123"

class Solution {
public: 
    string trim(const std::string&str){
        string nstr;
        int i=0;
        while(isspace(str[i])){
            i++;
        }
        
        for(;i<str.length();i++){
            if(isspace(str[i])){
                break;
            }
            nstr +=str[i];
        }
        return nstr;
    }
    
    int myAtoi(string str) {
        str = trim(str);
        if(str.length()==0 || (str[0]!='+'&&str[0]!='-'&& !isdigit(str[0]))){
            return 0;
        }
        
        int i=0;
        
        //consume sign char
        if(str[0] =='+' || str[0]=='-'){
            i++;
        }
        string nstr;
        while(isdigit(str[i])){
            nstr+=str[i];
            i++;
        }
        if(nstr.length()==0){
            return 0;
        }
        i=0;
        // consume meaningless zeros
        while(nstr[i]=='0'){
            i++;
        }
        nstr = nstr.substr(i);
        long long result = 0L;
        unsigned long long exp = 1; 
        for(int k=nstr.length()-1;k>=0;k--){
            result += ((int)(nstr[k]-'0'))*exp;
            
            if(exp> numeric_limits<int>::max()){
                return str[0]=='-'?numeric_limits<int>::min():numeric_limits<int>::max();
            }
            exp*=10;
            if(result> numeric_limits<int>::max()){
                return str[0]=='-'?numeric_limits<int>::min():numeric_limits<int>::max();
            }
        }
        return str[0]=='-'?-result:result;
    }
};