简介
Implement atoi which converts a string to an integer.
The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.
The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.
If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.
If no valid conversion could be performed, a zero value is returned.
Note:
- Only the space character
' 'is considered as whitespace character. - Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. If the numerical value is out of the range of representable values, INT_MAX (231 − 1) or INT_MIN (−231) is returned.
Example 1:
Input: "42" Output: 42
Example 2:
Input: " -42" Output: -42 Explanation: The first non-whitespace character is '-', which is the minus sign. Then take as many numerical digits as possible, which gets 42.
Example 3:
Input: "4193 with words" Output: 4193 Explanation: Conversion stops at digit '3' as the next character is not a numerical digit.
Example 4:
Input: "words and 987" Output: 0 Explanation: The first non-whitespace character is 'w', which is not a numerical digit or a +/- sign. Therefore no valid conversion could be performed.
Example 5:
Input: "-91283472332" Output: -2147483648 Explanation: The number "-91283472332" is out of the range of a 32-bit signed integer. Thefore INT_MIN (−231) is returned.
解题思路
1.首先把字符串前面的连续空格去掉;
2.判断下一个字符是否为+或-，若是跳3；若为数字，跳4；若为非数字非±，则直接返回0；
3.再判断+或-的后一个字符是否为数字，若为跳4；若不为直接返回0；
4.添加到新的字符串ss,判断该数字后面的是否为数字，若为重复4；若不为跳5；
5.得到的新字符串是由数字构成的字符串，再将其转换为整型即可。
代码
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
#define INT_MAX 2147483647
#define INT_MIN -2147483648
class Solution {
public:
int myAtoi(string str) {
bool flag = true; //判断正负
int n;
int j = 0;
string ss;
while(1) //跳过前面的空格
{
if(str[j] != ' ')break;
j++;
}
//数字前有负号
if(str[j] == '-' && str[j+1] >='0' && str[j+1] <='9')
{
flag = false;
}
//数字前有正号
else if(str[j] == '+' && str[j+1] >='0' && str[j+1] <='9')
{
flag = true;
}
//数字前出现其他符号
else if(str[j] < '0' || str[j] > '9')
{
return 0;
}
for(int i = j; i < str.length(); i++)
{
if(str[i] >= '0' && str[i] <= '9')
{
if(str[i+1] < '0' || str[i+1] > '9')
{
ss += str[i];
break;
}
ss += str[i];
}
}
if(flag == false)
{
ss = '-' + ss;
}
n = StringToInt(ss);
if(n > INT_MAX)
{
return INT_MAX;
}
if(n < INT_MIN)
{
return INT_MIN;
}
return n;
}
private:
int StringToInt(string str) //字符串转换为正型
{
stringstream ss;
ss<<str;
int i;
ss>>i;
return i;
}
};
int main()
{
string ss;
Solution su;
getline(cin, ss);
cout << su.myAtoi(ss);
return 0;
}
总结
1.string类型如果要把空格输入进去，需要使用getline(cin, str)
2.类型转换可以使用stringstream类