LeetCode第7題: 整數反轉

1. 本題比較簡單, 利用地板除和取模即可得出, 最後需要判斷是否溢位

class Solution:
    def reverse(self, x: int) -> int:
        i = 1
        if x < 0:
            i = -1
            x = -x
        mid = 0
        while x > 0:
            mid = mid * 10 + x % 10
            x //= 10
        if mid < -(1 << 31) or mid > ((1 << 31) - 1) :
            return 0
        return i * mid
 

LeetCode第8題: 字串轉換整數

1. 使用strip()去掉空格, 如果是空字元返回0
2. 使用分支結構逐層判斷即可

class Solution:
    def myAtoi(self, s: str) -> int:
        s = s.strip()
        if s == "":
            return 0
        k = 0
        flag = 0
        if s[0] in {'+', '-', '0', '1', '2', '3', '4', '5', '6', '7', '8', '9'}:
            if
 
 s[0] >= '0' and s[0] <= '9':
                k = 1
            else:
                k = 1
                flag = 1
            for i in range(1, len(s)):
                if (s[i] >= '0' and s[i] <= '9') or s[i] == '.':
                    k += 1
                else:
                    break
 
;
        if (k <= 0) or (flag == 1 and k <= 1):
            return 0
        i = int(float(s[0:k]))
        if i < -(1 << 31):
             return -(1 << 31)
        if i > ((1 << 31) -1):
             return ((1 << 31) - 1)
        return i

LeetCode第9題: 迴文數

1. 如果是負數直接返回False
2. 如果是正數, 反轉比較即可

class Solution:
    def isPalindrome(self, x: int) -> bool:
        if x < 0:
            return False
        return int(str(x)[::-1]) == x