链表

反转链表（力扣70）

思路1：转字符串之后判断

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

思路2：通过数字判断(这里要注意x发生变化了，最后判断时需要与原始的x进行比较)

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class Solution:
    def isPalindrome(self, x: int) -> bool:
        before = x
        if x < 0:
            return False
        res = 0
        while x>0:
            res = res*10 + x%10
            x = x//10
        return res == before

最小花费爬楼梯（力扣746）

思路：字典判断个数即可,小于等于1个则返回True

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from collections import Counter
class Solution:
    def canPermutePalindrome(self, s: str) -> bool:
        temp = Counter(s)
        count = 0
        for i in temp:
            if temp[i]%2 == 1:
                count += 1
        return count <= 1

参考