2021-01-12

007 整数反转

给出一个 32 位的有符号整数，你需要将这个整数中每位上的数字进行反转。

class Solution:
    def reverse(self, x: int) -> int:
        s = str(x)
        if s[0]=='-':
            x = int('-'+s[1:][::-1])
        else:
            x = int(s[::-1])

        return x if -2**31<x<2**31-1 else 0

008 字符串转换整数

class Solution:
    def myAtoi(self, str: str) -> int:
        i=0
        n=len(str)
        while i<n and str[i]==' ':
            i=i+1
        if n==0 or i==n:
            return 0
        flag=1
        if str[i]=='-':
            flag=-1
        if str[i]=='+' or str[i]=='-':
            i=i+1
        INT_MAX=2**31-1
        INT_MIN=-2**31
        ans=0
        while i<n and '0'<=str[i]<='9':
            ans=ans*10+int(str[i])-int('0')
            i+=1
            if(ans-1>INT_MAX):
                break
        ans=ans*flag
        if ans>INT_MAX:
            return INT_MAX
        return INT_MIN if ans<INT_MIN else ans

009 回文数

class Solution:
    def isPalindrome(self, x: int) -> bool:
        s = str(x)
        i,j = 0,len(s)-1

        if s[0] =='-':
            return False
        while i<= j:
            if s[i]!=s[j]:
                return False
        i += 1
        j -= 1
        return True