一行，不用多说

class Solution:
    def findDifference(self, nums1: List[int], nums2: List[int]) -> List[List[int]]:
        return[list(set(nums1)-set(nums2)),list(set(nums2)-set(nums1))]

正向遍历即可，我写的四行解法👇

class Solution:
    def minDeletion(self, nums: List[int]) -> int:
        res, n, flag = 0, len(nums), 0
        for i, v in enumerate(nums):
            if not flag or v != pre:pre,res,flag = v,res+1,1-flag
        return n-res+1 if res % 2 else n-res

六弦爷一行解法👇

class Solution:
    def minDeletion(self, nums: List[int], c = 0) -> int:
        return len(nums) - reduce(lambda r, x: (r[0] + 1, x) if r[0] & 1 == 0 or r[1] != x else r, nums, (0, -1))[0] // 2 * 2

先构造左边的回文数再反转，二行解法👇

class Solution:
    def kthPalindrome(self, queries: List[int], intLength: int) -> List[int]:
        half,base = (intLength+1) // 2,10** ((intLength + 1) // 2 - 1)
        return [(func:=lambda q:-1 if len(left:= str(base + q - 1)) > half else int(left + .join(left[i] for i in range(len(left) - 1 - intLength % 2, -1, -1))))(q) for q in queries]

动态规划，c++六行解法

class Solution {
          
   
public:
    int maxValueOfCoins(vector<vector<int>>& piles, int k) {
          
   
        vector<int> dp(k + 1);
        for (auto &p: piles)
            for (int i = k; i >= 0; i--) 
                for (int j = 1, sum = 0; j <= p.size() && i >= j; j++) 
                    {
          
   sum += p[j - 1];dp[i] = max(dp[i], dp[i - j] + sum);}
        return dp[k];
    }

};

总结

T2有坑，T4用python可能TLE。。。 T1+T2+T3+T4共1+1+2+6=10行代码，完成【20行完成周赛】的目标！