LeetCode 8. String to Integer (atoi)
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.
Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button to reset your code definition.
解析：
将给定的字符串转换为整数，功能类似于atoi。
因为输入时字符串，所以需要考虑的情况比较多。具体说来：
1.字符串可能以空格开始，需要从第一个非空字符开始计算。
2.允许数字以正负号开头。
3.遇到非法字符便停止转换，返回当前已经转换的值，如果开头就是非法字符则返回0。
4.转换的整数越界，停止处理返回0。
具体代码如下：
[cc lang=”C++”]
class Solution {
public:
int myAtoi(string str) {
int length = str.length();
long result = 0;
int flag = 1;
if(length <= 0 )
return 0;
int index = str.find_first_not_of(' ');//第一个非空字符开始
for(int i=index; i < length; i++)
{
if(str[i] == '-')
{
if(i+1
return INT_MAX;
if(result*flag < INT_MIN)
return INT_MIN;
}else{
break;
}
}
return result * flag;
}
};
[/cc]
update in 2019.5.25
另一个版本的代码，代码看上去更加简洁。
[cc lang="C++"]
class Solution {
public:
int myAtoi(string str) {
int sign = 1, base = 0, i = 0;
while (str[i] == ' ') { i++; }
if (str[i] == '-' || str[i] == '+') {
sign = 1 - 2 * (str[i++] == '-');
}
while (str[i] >= ‘0’ && str[i] <= '9') {
if (base > INT_MAX / 10 || (base == INT_MAX / 10 && str[i] – ‘0’ > 7)) {
if (sign == 1) return INT_MAX;
else return INT_MIN;
}
base = 10 * base + (str[i++] – ‘0’);
}
return base * sign;
}
};
[/cc]
主要解释一下条件里面的 str[i] – ‘0’ > 7
因为INT_MAX= 2147483647，最后一位数是7，如果超过7则会出现溢出的情况。
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