LeetCode(java)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.

题目描述：将字符串这边为整型

解法一：使用正则表达式匹配：

效率不高，leetcode通过时间为：48ms

public int myAtoi(String str) {
str = str.trim();
if(str.matches("[\\+|-]\\p{Blank}+\\d*"))
return 0;
boolean flag = false;
if(str.startsWith("-")||str.startsWith("+"))
{
if(str.startsWith("-"))
flag = true;
str = str.substring(1);
}
if(str.startsWith("-")||str.startsWith("+"))
return 0;
Pattern p = Pattern.compile("^(\\d+)");
Matcher m = p.matcher(str);
try {
if(m.find())
return flag?Integer.valueOf("-"+m.group()):Integer.valueOf(m.group());
} catch (NumberFormatException e) {
if(flag)
return Integer.MIN_VALUE;
else
return Integer.MAX_VALUE;
}
return 0;
}

解法二：从头开始遍历字符串：

时间复杂度：O(n)，leetcode通过时间为：3ms

public int myAtoi(String str) {
str = str.trim();//消去前后空格
int len = str.length();
if(len<1)
return 0;
long rs = 0;
boolean isNeg = false;
char[] arr = str.toCharArray();
int begin = 0;
if(arr[0] == '-')
{
isNeg = true;
begin++;
}
if(arr[0] == '+')
begin++;
for(int i=begin;i<len;i++)
{
if(arr[i]<'0' || arr[i]>'9')
break;
rs = rs*10+(arr[i]-'0');
if(rs>Integer.MAX_VALUE) //是否有益处可能
{
if(!isNeg)//正数则溢出
return Integer.MAX_VALUE;
else if(rs>-Integer.MIN_VALUE)//负数的话再判断是否溢出
return Integer.MIN_VALUE;
}
}
if(isNeg)
return (int)(-rs);
else
return (int)rs;
}