LeetCode08, string to integer (atoi)

topic

prompt:

The blank characters in this question only include the space character ''.
Assuming that our environment can only store 32-bit signed integers, the value range is [−231, 231 − 1]. If the value exceeds this range, please return INT_MAX (231 − 1) or INT_MIN (−231).

Example 1:

Input: "42"
Output: 42
Example 2:

Input: "-42"
Output: -42
Explanation: The first non-blank character is'- ', which is a negative sign.
We try our best to combine the minus sign with all subsequent numbers, and finally get -42.
Example 3:

Input: "4193 with words"
Output: 4193
Explanation: The conversion ends at the number '3' because the next character is not a number.
Example 4:

Input: "words and 987"
Output: 0
Explanation: The first non-blank character is'w', but it is not a number or a positive or negative sign.
Therefore, effective conversion cannot be performed.
Example 5:

Input: "-91283472332"
Output: -2147483648
Explanation: The number "-91283472332" exceeds the range of a 32-bit signed integer.
Therefore INT_MIN (−2 ³¹ ) is returned .
Passed 195,723 Submitted 938,532

1. Direct thinking-string regular matching

class Solution {
    
    
    public int myAtoi(String str) {
    
    
       str=str.stripLeading();//去除首部空格
        java.util.regex.Pattern pattern = java.util.regex.Pattern.compile("^[+|-]?\\d+");//\是转义字符，需要\\表示
        java.util.regex.Matcher matcher = pattern.matcher(str);
        //System.out.println(str.matches("^[+|-]?\\d+"));
        if(matcher.lookingAt()){
    
    //匹配成功,lookingAt()部分匹配，matches()是全部匹配
            str = str.substring(0,matcher.end());
            int res;
            int Limit = Integer.MAX_VALUE;
            try{
    
    
                if(str.charAt(0)=='-'){
    
    //是负数，可能越界值为 Integer.MIN_VALUE
                    Limit = Integer.MIN_VALUE;
                }//否则为正数
                res = Integer.parseInt(str);

            }catch (NumberFormatException e){
    
    
                res = Limit;
            }
            return res;
        }else
            return 0;



    }
}

Once accept, the efficiency is touching. It is estimated that the problem of quoting the class library.

2. If you have to make your own wheels

class Solution {
    
    
    public int myAtoi(String str) {
    
    
         //下面自己造轮子
        char[] chars = str.toCharArray();
        int start=-1;
        int flag=1;//判断是否为正数
        for (int i = 0; i < chars.length; i++) {
    
    //最多判断n次，就是全部是空格
            if(chars[i]==' '){
    
    //空格忽略
                continue;
            }
            if(chars[i]=='+'){
    
    //整数
                start = i;
                break;
            }
            if(chars[i]=='-'){
    
    //负数
                start = i;
                flag= -1;//只有这里可能是负数
                break;
            }
            if(!Character.isDigit(chars[i])){
    
    //不是数字或者+，-号开头则直接退出
                return 0;
            }
            if(Character.isDigit(chars[i])){
    
    //数字开头
                start = i;
                break;
            }
        }
        if(start<0)
            return 0;//全是空格
        if(chars[start]=='-'||chars[start]=='+') {
    
    
            start = start+1;
        }
        //也就是现在的start要么指向的是数字，要么指向的是符合

        int res=0;
        for(int i=start;i<chars.length;i++){
    
    
           if(Character.isDigit(chars[i])){
    
    //是数字则可以进行转化
               int pop = flag*(chars[i]-'0');
               //判断是否越界
               if (res > Integer.MAX_VALUE/10 || (res == Integer.MAX_VALUE / 10 && pop > 7))
                   return Integer.MAX_VALUE;
            
               if (res < Integer.MIN_VALUE/10 || (res == Integer.MIN_VALUE / 10 && pop < -8)) 
                   return Integer.MIN_VALUE;
               res = res*10+pop;//需要放后边，因为我们是先判断/10合要求了，那么它*10才会符合要求
           }else
               break;
        }
        return res;

    }
}

As expected: cowhide!