ascii to integer的实现。判断的方法是进位累加。要格外注意'1'和1这样的char和int的区别。
- 有效字符后面的省略。 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.
- 首次出现的sequence不是有效数字，返回0。
- str为空或者全是空格，返回0。
- outOfRange的时候返回INT_MIN/INT_MAX
public int myAtoi(String str) {
int total = 0, sign = 1, index = 0;
if (str == null || str.length() == 0) return 0;
//jump through spaces
while (index < str.length() && str.charAt(index) == ' ') index++;
//或者这么写：if(total > (Integer.MAX_VALUE - add) / 10)
if (str.charAt(index) == '+' || str.charAt(index) == '-') {
sign = str.charAt(index) == '+' ? 1 : -1;
index++;
}
//计算方法是进位累加
while (index < str.length()) {
//易错
int digit = str.charAt(index) - '0';
if (digit < 0 || digit > 9) break;
if (total > Integer.MAX_VALUE / 10 || total == Integer.MAX_VALUE / 10 && digit > Integer.MAX_VALUE % 10) {
total = sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
break;
}
total = total * 10 + digit;
index++;
}
return total * sign;
}
注意INTMAX和INTMIN分别是2147483647和-2147483648，并不需要分开判断。很巧妙。
ref:
https://discuss.leetcode.com/topic/12473/java-solution-with-4-steps-explanations
