思路：模拟，排除边界情况

Python：

class Solution:
    def strToInt(self, str: str) -> int:
        if not str or (len(str)==1 and (str[0]<'0' or str[0]>'9')):
            return 0
        str=str.strip()
        count=len(str)
        digit_flag=False
        for i in range(len(str)):
            if str[i]=='-' or str[i]=='+':
                if not digit_flag:
                    continue
                else:
                    count=i
                    break
            elif (str[i]>='0' and str[i]<='9'):
                digit_flag=True
                continue
            else:
                count=i
                break
        try:
            if str[0] == '-':
                if str[1]>'9' or str[1]<'0':
                    return 0
                try:
                    tmp_int=int(str[1:count])
                    res=0-tmp_int
                except:
                    tmp_int=int(str[1:])
                    res=0-tmp_int
            elif str[0]=='+':
                if str[1]>'9' or str[1]<'0':
                    return 0
                try:
                    res=int(str[1:count])
                except:
                    res=int(str[1:])
            elif str[0]>='0' and str[0]<='9':
                try:
                    tmp_int=int(str[:count])
                    res=tmp_int
                except:
                    tmp_int=int(float(str))
                    res=tmp_int
            else:
                return 0
        except:
            return 0
        if res>2**31-1:
            return 2**31-1
        elif res<(-2)**31:
            return (-2)**31
        else:
            return res

标签：转换成,elif,return,Offer,digit,len,flag,str,67
来源： https://www.cnblogs.com/tiuyoaix/p/16171935.html