題目描述
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.
我的解答是:public static int myAtoi(String str) {
if (str == null || str.length() == 0)
return 0;//
str = str.trim();
char firstChar = str.charAt(0);
int sign = 1, start = 0, len = str.length();
int sum = 0;
if (firstChar == '+') {
sign = 1;
start++;
} else if (firstChar == '-') {
sign = -1;
start++;
}
if(start==len)return 0;
try {
return sign*Integer.parseInt(str.substring(start));
} catch (NumberFormatException e) {
// TODO Auto-generated catch block
return 0;
}
}
提交時顯示錯誤:
https://img-blog.csdn.net/20170327235301435?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQvaWNhaTg4OA==/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/Center而leetcode給出的答案:
public static int myAtoi(String str){
if (str == null || str.length() == 0)
return 0;//
str = str.trim();
char firstChar = str.charAt(0);
int sign = 1, start = 0, len = str.length();
long sum = 0;
if (firstChar == '+') {
sign = 1;
start++;
} else if (firstChar == '-') {
sign = -1;
start++;
}
for (int i = start; i < len; i++) {
if (!Character.isDigit(str.charAt(i)))
return (int) sum * sign;
sum = sum * 10 + str.charAt(i) - '0';
if (sign == 1 && sum > Integer.MAX_VALUE)
return Integer.MAX_VALUE;
if (sign == -1 && (-1) * sum < Integer.MIN_VALUE)
return Integer.MIN_VALUE;
}
return (int) sum * sign;
}
測試:"+-2",輸出爲0,而提交時卻ac了，實在想不明白