[leetcode]String to Integer (atoi)

String to Integer (atoi)

 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.

算法思路：

坑比较多，没难度。开整之前建议先把str预处理一下

【注意】边界case：

全空格、全非法字符、空串<-这三种预处理之后都变成空串了。

溢出。因此先把result声明成long型。

符号处理：

两个boolean变量表示正负号，但是一旦有了某个符号，第二个符号就变成非法字符了。字符串中未出现符号，按照正号处理。

代码如下：

public class Solution {
    public int atoi(String str) {
        if(str == null || str.trim().length() == 0) return 0;
        str = str.trim();
        StringBuilder sb = new StringBuilder();
        int length = str.length();
        ////preprocess,only maintain num and operater
        for(int i = 0; i < length; i++){
            if(str.charAt(i) == '+' || str.charAt(i) == '-'  || (str.charAt(i) <= '9' && str.charAt(i) >= '0')){
                sb.append(str.charAt(i));
            }else break;
        }
        length = sb.length();
        boolean positive = false,negative = false;
        long res = 0;
        for(int i = 0; i < length; i++){
            //the second time appearrance of operater is invalide
            if((sb.charAt(i) == '+' || sb.charAt(i) == '-') && (positive || negative)) 
                break;
            if(sb.charAt(i) == '+'){
                positive = true;
            }else if(sb.charAt(i) == '-'){
                negative = true;
            }else{
                res = res * 10 + (sb.charAt(i) - '0');
            }
        }
        if(!positive && !negative) positive = true;
        // process overflow situation
        if(positive) return res > Integer.MAX_VALUE ? Integer.MAX_VALUE : (int) res;
        return res > Integer.MAX_VALUE ? Integer.MIN_VALUE : (int)res * -1;        
    }
}