String integer conversion (atoi)
Atoi you to implement a function, it can convert a string to an integer.
First, the function will begin with a space character discard useless if necessary, until the find to the first non-space character so far.
When we find the first non-space character is a positive or negative number, the combination of the symbols as much as possible with consecutive numbers up later, as the sign of integer; if the first non-space character is figures, which directly after the continuous numeric characters are combined to form an integer.
In addition to the string after a valid integer part may also exist extra characters, these characters can be ignored, they should not affect a function.
Note: if the character string in the first non-space character is not a valid integer character string is empty or contains only white space character string, then you will not need to be a function of conversion.
In any case, if the function can not effectively convert, 0 is returned.
Description:
We assume that the size of the environment can store 32-bit signed integer, then the value range of [-2 ^ 31, 2 ^ 31 - 1] If the value exceeds this range, qing return INT_MAX (2 ^ 31 - 1) or INT_MIN (-2 ^ 31).
Example 1:
Input: "42"
Output: 42
Example Two:
Input: "-42"
Output: -42
Explanation: a first non-blank character '-', it is a negative sign.
We will all digital consecutive negative number and later combined as much as possible, and finally get -42.
Example Three:
Input: "4193 with words"
Output: 4193
Explanation: converting the digital OFF '3', because the next character is not numeric.
Example Four:
Input: "words and 987"
Output: 0
Explanation: a first non-blank character 'w', but it is not a positive number or negative number.
Therefore, the conversion can not be performed effectively.
Example five:
Input: "-91283472332"
Output: -2147483648
explanation: the number "-91283472332" Over the range of 32-bit signed integer.
Accordingly return INT_MIN (-2 ^ 31).
Algorithms ideas:
The above-described character string entered slowly filtration conditions, ultimately to give a string of numeric strings strNum, wherein the orientation of the string 10 within. (2 ^ 31 = 2,147,483,647), then either output.
- First, determine the null character, directly back to 0.
if(str.length() < 1)
- Filtration beginning character is empty, get new strings.
int iNullStrCnt = 0;
while(iNullStrCnt < str.length() && str[iNullStrCnt] == ' '){
iNullStrCnt++;
}
str = str.substr(iNullStrCnt,str.length());
- Determining a first non-blank character, or if it is not a digital '+', '-', the process directly returns 0
if yes, place '+', '-', at the beginning of the symbol to obtain a digital string.
if(str[0] < '0' || str[0] > '9'){
if(str[0] == '-'){
bSybolFlag = -1;
}
else if(str[0] == '+'){
bSybolFlag = 1;
}
else{
return 0;
}
str = str.substr(1,str.length());
- Write function to get the string of numbers. Requirements: Get number of consecutive, when there is non-numeric, it represents the end of the data, returns the number of consecutive preceding string. Such as: 123a123 - 123 returns.
string getNumStr(string str){
string strNum = "";
bool bFlag = true;
for(int i = 0; i < str.length(); i++)
{
if(str[i] >= '0' && str[i] <= '9'){
//跳过，开始字符为0
if(bFlag && str[i] == '0')
{
continue;
}
bFlag = false;
strNum.push_back(str[i]);
}
else{
break;//非数字，跳出本次循环
}
}
return strNum;
}
- After obtaining the number string, and accordingly beyond the process are:
(1) numeric string more than 10, 0 is returned.
(2) The calculated result exceeds the maximum value of return INT_MAX INT_MAX.
(3) The calculated result is smaller than the maximum value of return INT_MIN INT_MIN.
Code:
class Solution {
public:
int myAtoi(string str) {
//空字符 直接返回0
if(str.length() < 1)
return 0;
//过滤开头字符为空
int iNullStrCnt = 0;
while(iNullStrCnt < str.length() && str[iNullStrCnt] == ' '){
iNullStrCnt++;
}
str = str.substr(iNullStrCnt,str.length());
cout << "去除空格字符str ="<<str<<endl;
//判断第一个非空字符的符号
int bSybolFlag = 1;
string strNum = "";
//判断第一个字符是否为‘+’、‘-’，数字，否则直接返回0
//开头字符非数字的情况
if(str[0] < '0' || str[0] > '9'){
if(str[0] == '-'){
bSybolFlag = -1;
}
else if(str[0] == '+'){
bSybolFlag = 1;
}
//非‘+’ 非‘-’ 直接返回0
else{
cout<<"非‘+’ 非‘-’ 直接返回0"<<endl;
return 0;
}
str = str.substr(1,str.length());
cout << "去除符号字符str ="<<str<<endl;
strNum = getNumStr(str);
}
else{
strNum = getNumStr(str);
}
cout << "得到数字字符串str ="<<strNum<<"bsybolflag = "<<bSybolFlag<<"最大值="<<INT_MAX<<"最小值 = "<<INT_MIN<<endl;
//数据超出long的情况直接丢掉
if(strNum.length()>10){
if(bSybolFlag == 1){
return INT_MAX;
}
else if(bSybolFlag == -1){
return INT_MIN;
}
}
//结果
long lResultNum = 0;
for(int i = 0;i < strNum.length(); i++)
{
lResultNum = lResultNum*10 + strNum[i] - '0';
if(lResultNum*bSybolFlag < INT_MIN) return INT_MIN;
if(lResultNum*bSybolFlag > INT_MAX) return INT_MAX;
}
return lResultNum*bSybolFlag;
}
string getNumStr(string str){
string strNum = "";
bool bFlag = true;
for(int i = 0; i < str.length(); i++)
{
if(str[i] >= '0' && str[i] <= '9'){
//跳过，开始字符为0
if(bFlag && str[i] == '0')
{
continue;
}
bFlag = false;
strNum.push_back(str[i]);
}
else{
break;//非数字，跳出本次循环
}
}
return strNum;
}
};
