题目
题目很简单，就是写一个函数把string转换成int，但是通过率只有可怜的11%，难点是要考虑所有情况，特别是int溢出边界，反正我是写了2个小时还没解决，先放到这，有空接着搞，现在应该还有最后一个bug。
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.
代码
public class Solution {
public int atoi(String str) {
String num_str="";
char[] str_char=str.toCharArray();
char[] sympo="-+".toCharArray();
boolean abs=true;
for(int i=0;i<str.length();i++){
if(str_char[i]==sympo[0] ||str_char[i]==sympo[1]){
if(!Character.isDigit(str_char[i+1])){
return 0;
}
if(str_char[i]==sympo[0]){
abs=false;
}
}
if(Character.isDigit(str_char[i])){
num_str+=String.valueOf(str_char[i]);
}
else{
if(num_str.length()!=0){
break;
}
}
// if(Character.isDigit(str_char[i])){
// num_str+=String.valueOf(str_char[i]);
// }
if(num_str!=""){
if(Math.abs(Integer.parseInt(num_str))>=Integer.MAX_VALUE / 10){
return Integer.MAX_VALUE;
}
}
}
if(num_str!=""){
if(abs){
return Integer.parseInt(num_str);
}
else{
return -Integer.parseInt(num_str);
}
}
else{
return 0;
}
}
}
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