String to Integer (atoi) leetcode

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.

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.

这道题要考虑 1. 越界问题. 2. 正负号问题 3. 空格问题(用string.trim()) 4.异常符号出现(符号前面保留 后面全删除)

public class Solution {
    public int myAtoi(String str) {
        if(str == null) {
            return 0;
        }
        str = str.trim();
        if (str.length() == 0) {
            return 0;
        }
        int index = 0;
        int tem = 1;
        long result = 0;//因为result可能大于 int的范围
        if (str.charAt(index) == '-') {
            tem = -1;
            index++;
        } else if (str.charAt(index) == '+') {
            index++;
        }
        for (; index < str.length(); index++) {
            if (str.charAt(index) < '0' || str.charAt(index) > '9') {
                break;
            }
            int mid = str.charAt(index) - '0';
            result = result * 10 + mid;
            if (result > Integer.MAX_VALUE) {
                break;
            }
        }
        if (result * tem >  Integer.MAX_VALUE) {
            return Integer.MAX_VALUE;
        }
        if (result * tem < Integer.MIN_VALUE) {
            return Integer.MIN_VALUE;
        }
        return (int)result * tem;//因为result为long 但是最后要返回int
    }
}