61(groupby,sum和if结合的计算) 62(left,substr,字符串的截取和大小写), 70.行转列(groupby+sumif或者groupby+case when或者left join穷举) 72.列转行 用union all 75.date_format参数 78.TIMESTAMPDIFF、UNIX_TIMESTAMP 和 ABS的结合使用 63~69,71,74,77~
Can you solve this real interview question? Find the Duplicate Number - Given an array of integers nums containing n + 1 integers where each integer is in the range [1, n] inclusive. There is only one repeated number in nums, return this repeated number
Leetcode 399 Evaluate Division Leetcode 1274 Number of Ships in a Rectangle Leetcode 1376 Time Needed to Inform All Employees Leetcode 694 Number of Distinct Islands Leetcode 131 Palindrome Partitioning 基于排列组合的DFS: 其实与图类DFS方法一致,但是排列组合的特征更明显 Leetcode 17 Letter Combinations...
codeforces_solutionsProblemSolution 1A - Theatre Square Rust 4A - Watermelon Rust 71A - Way Too Long Words Rust 158A - Next Round Rust 231A - Team Rustleetcode_solutions(problem number with 🔒 suffix need leetcode VIP to unlock)#ProblemSolutionsCategory/Comment...
First off, we can easily show that the constraints of the problem imply that a cycle must exist. Because each number in nums is between 11 and nn, it will necessarily point to an index that exists. Therefore, the list can be traversed infinitely, which implies that there is a cycle. Ad...
2 3 4 5 6 7 8 9 10 11 12 13 14 classSolution(object): defmissingNumber(self, nums): """ :type nums: List[int] :rtype: int """ if0notinnums: return0 m=max(nums) s=sum(nums) ms=(m+1)*m/2 ifms==s: returnm+1 ...
Note: All explanations are written in Github Issues, please do not create any new issue or pull request in this project since the problem index should be consistent with the issue index, thanks! ('$' means the problem is locked on Leetcode, '*' means the problem is related to Database...
编写一个算法来判断一个数 n 是不是快乐数。 「快乐数」定义为: 对于一个正整数,每一次将该数替换为它每个位置上的数字的平方和。然后重复这个过程直到这个数变为 1,也可能是 无限循环 但始终变不到 1。如果 可以变为 1,那么这个数就是快乐数。如果 n 是快乐数就返回 true ;不是,则返回 false 。
problem 268. Missing Number solution1:等差数列求和公式 根据题意,(0, 1, 2,...,n),共有(n+1)个数,丢失了一个数之后,剩余的数组成的数组,求解丢失的数据。 等差数列求和减去数组累加和得到的结果即为所求数据。 class Solution { ...
Note:It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one. However, here is a list of characters that can be in a valid decimal number: Numbers 0-9 Exponent - "e" ...