leetcode-8-字符串转换整数（atoi)

题目描述：

方法一：正则

import re
class Solution:
    def myAtoi(self, str: str) -> int:
        INT_MAX = 2149483647
        INT_MIN = -2147483648
        num = re.findall(r'^[\+\-]?\d+',str.lstrip())
        return max(min(int(*num),INT_MAX),INT_MIN)

class Solution:
    def myAtoi(self, str: str) -> int:
        return max(min(int(*re.findall('^[\+\-]?\d+', str.lstrip())), 2**31 - 1), -2**31)

java版：

package test;
 
import java.util.regex.Matcher;
import java.util.regex.Pattern;
 
class Solution{
    public int myAtoi(String str) {
        str = str.trim();
        Pattern p = Pattern.compile("^[\\+\\-]?\\d+");
        Matcher m = p.matcher(str);
        int value = 0;
        if(m.find()) {
            try {
                value = Integer.parseInt(str.substring(m.start(), m.end()));
            } catch (NumberFormatException e) {
                value = str.charAt(0) == '-'? Integer.MIN_VALUE:Integer.MAX_VALUE;
            }
 
        }
        return value;
    }
}
public class Test {
    public static void main(String[] args) {
          Solution s = new Solution();
          System.out.println(s.myAtoi(" -0456"));
    }
}

方法二：

利用双指针加eval()函数
思路：

用left=0，right标记第一个数字和最后一个数字出现的位置

首先strip()去掉左右的空格

然后判断第一个字符是不是正负号，是的话left置为1

判断left后一个字符是否为数字，不是直接返回0

遍历不断更新right，找到第一个不是数字的字符后break

用res=str[left:right+1]提取纯数字字符串

需要注意的是，可能数字全为0.所以用lstrip()将左边的0清除

最后利用eval()函数将双引号去除，并判断是否需要填正负号

注意一些细节就ok了

下面代码都写有注释，用时28ms时间复杂度o(n)

class Solution:
    def myAtoi(self, str: str) -> int:
        str = str.strip()
        if str == "":
            return 0
        left = 0
        right = 0
        maxi=2147483647
        mini=-2147483648
        if str[0]=="+" or str[0]=="-":
            left = 1
        if(left==1 and len(str)==1) or str[left]<"" or str[right]>"":
            return 0
        for i in range(left,len(str)):
            if str[i]>='' and str[i]<='':
                right = i
            else:
                break
        res = str[left:right+1].lstrip("")
        if len(res)==0:
            return 0
        else :
            res=eval(res)
        if left==1 and str[0]=='-':#判断正负
            res=-res
        if res>maxi:
            return maxi
        elif res<mini:
            return mini
        else :
            return res
                

方法三：自动机

INT_MAX = 2147483647
INT_MIN = -2147483648
 
class Automaton:
    def __init__(self):
        self.state = 'start'
        self.sign = 1
        self.ans = 0
        self.table = {
            'start':['start','signed','in_number','end'],
            'signed': ['end', 'end', 'in_number', 'end'],
            'in_number': ['end', 'end', 'in_number', 'end'],
            'end': ['end', 'end', 'end', 'end'],
        }
 
    def get_col(self,c):
        if c.isspace():
            return 0
        if c == '+' or c == '-':
            return 1
        if c.isdigit():
            return 2
        return 3
    def get(self,c):
        self.state = self.table[self.state][self.get_col(c)]
        if self.state == 'in_number':
            self.ans = self.ans*10 + int(c)
            self.ans = min(self.ans,INT_MAX) if self.sign == 1 else min(self.ans,-INT_MIN)
        elif self.state == 'signed':
            self.sign = 1 if c == '+' else -1
class Solution:
    def myAtoi(self, str: str) -> int:
        automaton = Automaton()
        for c in str:
            automaton.get(c)
        return automaton.sign * automaton.a

java版

class Solution {
 
    class Automaton {
    final String START = "start";
    final String SIGNED = "signed";
    final String IN_NUM = "in_number";
    final String END = "end";
    String state = START;
    Map<String, String[]> map;
    public int sign = 1;
    public long ans = 0;
 
    public Automaton() {
        map = new HashMap<>();
        map.put(START, new String[]{START, SIGNED, IN_NUM, END});
        map.put(SIGNED, new String[]{END, END, IN_NUM, END});
        map.put(IN_NUM, new String[]{END, END, IN_NUM, END});
        map.put(END, new String[]{END, END, END, END});
    }
 
    public int get_col(char c) {
        if (c == ' ') return 0;
        if (c == '+' || c == '-') return 1;
        if (c >= '0' && c <= '9') return 2;
        return 3;
    }
 
    public void get(char c) {
        state = map.get(state)[get_col(c)];
        if (state.equals(IN_NUM)) {
            ans = ans * 10 + c - '0';
            if (sign == 1) {
                ans = Math.min(ans, Integer.MAX_VALUE);
            } else {
                // -(long)Integer.MIN_VALUE，这个操作有点东西，不然越界
                ans = Math.min(ans, -(long)Integer.MIN_VALUE);
            }
        } else if (state.equals(SIGNED))
            sign = c == '+' ? 1 : -1;
    }
    }
 
    public int myAtoi(String str) {
        Automaton automaton = new Automaton();
        char[] c = str.toCharArray();
        for (char ch : c) {
            automaton.get(ch);
        }
        return automaton.sign * ((int) automaton.ans);
    }
}