【JAVA、C++】 LeetCode 008 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.

解题思路：

本题难度并不大，比较无聊，主要是要考虑到几个边界条件，本人也是提交数次才通过的。

JAVA代码如下：

static public int myAtoi(String str) {
if (str == null )
return 0;
str = str.trim();
if(str.length()==0)
return 0;
boolean isNagetive = false;

if (str.charAt(0) == '-')
isNagetive = true;
long result = 0;
for (int i = 0; i < str.length(); i++) {
if(i==0&&(str.charAt(0)=='-'||str.charAt(0)=='+'))
continue;
int temp = str.charAt(i) - '0';
if (temp >= 0 && temp <= 9)
{
if(isNagetive){
result=result * 10 - temp;
}
else result = result * 10 + temp;
if (result > Integer.MAX_VALUE)
return Integer.MAX_VALUE;

if (result < Integer.MIN_VALUE)
return Integer.MIN_VALUE;
}

else break;
}
return (int) result;
}

C++

#include<algorithm>
using namespace std;
class Solution {
public:
int myAtoi(string str) {
bool isFirstChar = true, isNagetive = false;
long result = 0;
for (int i = 0; i < str.length(); i++) {
if (isFirstChar&&str[i] == ' ')
continue;
if (isFirstChar&&str[i] == '-') {
isNagetive = true;
isFirstChar = false;
continue;
}
if (isFirstChar&&str[i] == '+') {
isFirstChar = false;
continue;
}
int temp = str[i] - '0';
if (temp >= 0 && temp <= 9)
{
if (isNagetive) {
result = result * 10 - temp;
}
else result = result * 10 + temp;
if (result > INT_MAX)
return INT_MAX;

if (result < INT_MIN)
return INT_MIN;
}
else break;
isFirstChar = false;
}
return (int)result;
}
};