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):
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Requirements for atoi:
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. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.
翻译大概意思是：
1、先丢弃字符串前面的空白字符知道扫到第一个非空白字符；然后将一个合法的数值的字符串以及其前面的一个正负号一同装换为int。
2、该合法数值子字符串中或后，可能包括其它非数值的特殊字符，忽略这些子字符串；
3、若第一个非空子字符串非护法的整型数值。或根本就没有，甚至该字符串可能都是特殊字符或空字符，则不作转换0；
4、若没有有效地转换，则返回0，若有返回数值超出边界，则返回边界值。
一、思路：
二、Java程序
public class Solution {
public int myAtoi(String str) {
int sl = str.length();
int index=0;
char tempc; //获取字符
boolean flag = false; //负数标志
int firstI=-1;
long re=0;
if(sl<=0)
return 0;
//1. 開始至第一个非空字符
while(str.charAt(index)==' ')
{
index++;
if(index>=sl) //超过字符串长度
return 0;
}
//2. 扫到第一个非空字符,推断是否合法
tempc = str.charAt(index);
//2.1 若是+-或数值则进行，否则则返回0
if(tempc=='-'||tempc=='+'||(tempc<='9'&&tempc>='0'))
{
//2.2 若是数值则跳出推断
if((tempc<='9'&&tempc>='0'))
{
}else //2.3 若是+-号，则推断接下来的是否为数值
{
if(tempc=='-') flag = true;//2.3.1 若是-号。则设置标志位
index++; //2.4 若是+-号，则推断下一位是不是数值，是则合法，不是则返回0
tempc = str.charAt(index);
if((tempc<='9'&&tempc>='0'))
{}else
{
return 0;
}
}
}else //2.1 若是+-或数值则进行，否则则跳出
{
return 0;
}
//3. 開始扫描有效数字
while(index<sl)
{
tempc = str.charAt(index);
if(tempc<='9'&&tempc>='0')
{
re = re*10 + (str.charAt(index)-'0');
if(re>=Integer.MAX_VALUE||re<=Integer.MIN_VALUE)
break;
}else
{
break;
}
index++;
}
if(flag==true)
{ re=-re;
re = re<Integer.MIN_VALUE?Integer.MIN_VALUE:re;
}else
{
re = re>Integer.MAX_VALUE?Integer.MAX_VALUE:re;
}
return (int)re;
}
}