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.
思路分析:- 首先判断为空,返回0
- 考虑前面是空格,使用trim()去掉,然后判断长度是否为0,是的话,返回0
- 判断第一个字符是不是+和-,设置变量sign记录。
- 循环取得字符串的数字,考虑字符串中有非数字,遇到就退出,保留前面的数字
- 考虑溢出的情况,溢出返回Integer的最大值和最小值
public class Solution {
public int myAtoi(String str) {
//首先判断空值
if(str == null){
return 0;
}
//去掉空格的情况
str = str.trim();
if(str.length() == 0){
return 0;
}
int sign = 1;
int index = 0;
if(str.charAt(index) == '+'){
index++;
}else if(str.charAt(index) == '-'){
index++;
sign = -1;
}
//取得数字部分,遇到溢出和非数字退出
long number = 0;
for(;index < str.length();index++){
if(str.charAt(index) < '0' str.charAt(index) > '9'){
break;
}
number = number *10 + (str.charAt(index) - '0');
if(number >= Integer.MAX_VALUE){
break;
}
}
if(number * sign >= Integer.MAX_VALUE){
return Integer.MAX_VALUE;
}
if(number * sign <= Integer.MIN_VALUE){
return Integer.MIN_VALUE;
}
return (int)number*sign;
}
}