package com.jiucaiyuan.net.algrithm.sum;



/**

 * 将一个字符串转化为32位有符号整数，（类似C/C++中的atoi函数）

 *

 * @author jiucaiyuan on 2022/3/5.

 */

public class MyAtoi {

    public static int myatoi(String s) {

        char[] chars = s.toCharArray();

        int len = chars.length;

        //去掉空格的情况

        int index = 0;

        while (index < len && chars[index] == ' ') {

            index++;

        }

        // 排除极端情况，例如："    "

        if (index == len) {

            return 0;

        }



        // 设置符号

        int sign = 1;

        char firstChar = chars[index];

        if (firstChar == '-') {

            index++;

            sign = -1;

        } else if (firstChar == '+') {

            index++;

        }



        //last，用于记录上一次的res，以此来判断是否溢出的情况

        int res = 0, last = 0;

        while (index < len) {

            char c = chars[index];

            if (c < '0' || c > '9') {

                break;

            }

            int tem = c - '0';

            last = res;

            res = res * 10 + tem;



            //如果不相等就是溢出了

            if (last != res / 10) {

                return (sign == (-1)) ? Integer.MIN_VALUE : Integer.MAX_VALUE;

            }

            index++;

        }

        return res * sign;

    }



    public static void main(String[] args) {

        String source1 = "42";

        int target1 = myatoi(source1);

        System.out.println(source1 + "：" + target1);



        String source2 = "-42";

        int target2 = myatoi(source2);

        System.out.println(source2 + "：" + target2);



        String source3 = "4193 with words";

        int target3 = myatoi(source3);

        System.out.println(source3 + "：" + target3);

    }

}