把字符串转换成整数

 //leetcode8

public class Solution {

   public int myAtoi(String str) {

       int index = 0; 

       int sign = 1; 

       int total = 0;

      //1. Empty string 

       if(str.length() == 0) {

          return 0;

       }

       //2. Remove Spaces  去掉string开头处的空格

       while (str.charAt(index) == ' ' && index < str.length())

          index++;

       //3. Handle signs 如果string开头是+，最后结果为正

       if(str.charAt(index) == '+' || str.charAt(index) == '-') {

          sign = str.charAt(index) == '+' ? 1 : -1;

          index++; //要记住判断完后，index加1

       }

       //4. Convert number and avoid overflow

       while(index < str.length()) {

          int digit = str.charAt(index) - '0'; //记住要减'0'，否则str等于111，str.charAt(0)是1,1的ascii码是49 

          if(digit < 0 || digit > 9) 

             break;

       //check if total will be overflow after 10 times and add digit  判断是不是overflow，如果满足if条件，就会大于2^31-1，是的话就取到 Integer.MAX_VALUE

       if(Integer.MAX_VALUE/10 < total || Integer.MAX_VALUE / 10 == total && Integer.MAX_VALUE % 10 < digit)

          return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE;

       total = 10 * total + digit;

       index ++;

   }

   return total * sign;

   }

}