益智小游戏 Python 六

本文参考自力扣题目

1、9. 回文数

给你一个整数 x ，如果 x 是一个回文整数，返回 true ；否则，返回 false 。

回文数是指正序（从左向右）和倒序（从右向左）读都是一样的整数。例如，121 是回文，而 123 不是。

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

2、

class Solution:
    def romanToInt(self, s: str) -> int:
        sing = {
    
    'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000}
        dou = {
    
    'IV': 4,'IX':9, 'XL': 40, 'XC':90, 'CD':400, 'CM':900}
        
        index = 0
        num = 0
        while index < len(s):
            if s[index:index+2] in dou:
                num = num + dou[s[index:index+2]]
                index = index + 2
            else:
                num = num + sing[s[index:index+1]]
                index = index + 1
        return num

3 14. 最长公共前缀

class Solution:
    def longestCommonPrefix(self, strs: List[str]) -> str:
        if len(strs) == 0:
            return ""
        elif len(strs) == 1:
            return strs[0]
        elif len(strs) == 2:
            num = []
            a = strs[0]
            b = strs[1]
            lenth = min(len(a), len(b))
            for i in range(lenth):
                if a[i] == b[i]:
                    num.append(a[i])
                else: break
            return ''.join(num)

        else:
            num = []
            a = strs[0]
            b = strs[1]
            lenth1 = min(len(a), len(b))
            for i in range(lenth1):
                if a[i] == b[i]:
                    num.append(a[i])
                else: break

            for _ in range(2, len(strs)):
                c = strs[_]
                lenth2 = min(len(num), len(c))
                newnum = []
                for i in range(lenth2):
                    if num[i] == c[i]:
                        newnum.append(num[i])
                    else: break
                num = newnum
            return ''.join(newnum)

让代码变短！

class Solution:
    def longestCommonPrefix(self, strs: List[str]) -> str:
        if not strs:
            return ""
        
        length, count = len(strs[0]), len(strs)
        for i in range(length):
            c = strs[0][i]
            if any(i == len(strs[j]) or strs[j][i] != c for j in range(1, count)):
                return strs[0][:i]
        
        return strs[0]