leetcode刷题记录171-180 python版

前言

继续leetcode刷题生涯
这里记录的都是笔者觉得有点意思的做法
参考了好几位大佬的题解，尤其是powcai大佬和labuladong大佬，感谢各位大佬

171. Excel表列序号

# 26进制转10进制
class Solution:
    def titleToNumber(self, s: str) -> int:
        res = 0
        bit = 1
        for a in s[::-1]:
            res += (ord(a) - 64) * bit
            bit *= 26
        return res

172. 阶乘后的零

# 递归 数5的个数
class Solution:
    def trailingZeroes(self, n: int) -> int:
        return n // 5 + self.trailingZeroes(n // 5) if n != 0 else 0

173. 二叉搜索树迭代器

class BSTIterator:
    def __init__(self, root: TreeNode):
        self.stack = []
        self.push_stack(root)
    def next(self) -> int:
        """
        @return the next smallest number
        """
        tmp = self.stack.pop()
        if tmp.right:
            self.push_stack(tmp.right)
        return tmp.val
    def hasNext(self) -> bool:
        """
        @return whether we have a next smallest number
        """
        return bool(self.stack)
    def push_stack(self, node):
        while node:
            self.stack.append(node)
            node = node.left

174. 地下城游戏

# 动态规划
class Solution:
    def calculateMinimumHP(self, dungeon: List[List[int]]) -> int:
        row, col = len(dungeon), len(dungeon[0])
        dp = [[0] * col for _ in range(row)]
        dp[-1][-1] = max(1 - dungeon[-1][-1], 1)
        for i in range(col - 2, -1, -1):
            dp[-1][i] = max(1, dp[-1][i+1] - dungeon[-1][i])
        for i in range(row - 2, -1, -1):
            dp[i][-1] = max(1, dp[i+1][-1] - dungeon[i][-1])
        for i in range(row - 2, -1, -1):
            for j in range(col - 2, -1, -1):
                dp[i][j] = max(min(dp[i+1][j], dp[i][j+1]) - dungeon[i][j], 1)
        return dp[0][0]

175. 组合两个表

select FirstName, LastName, City, State
from Person left join Address
on Person.PersonId = Address.PersonId;

176. 第二高的薪水

# limit
select (
   select DISTINCT Salary
   from Employee
   order by Salary DESC
   limit 1,1)
as SecondHighestSalary;
# max
select max(Salary) as SecondHighestSalary
from Employee
where Salary < (select max(Salary) from Employee);

177. 第N高的薪水

CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
    # 定义变量
    declare p int;
    # 变量赋值
    set p=n-1;
RETURN (      
        select ifnull(
                        (
                        # LIMIT a OFFSET b 方法  
                        #select distinct salary from employee order by salary desc limit 1 OFFSET P
                        # LIMIT a,b 方法
                        select distinct salary from employee order by salary desc limit P,1
                        ),null) as SecondHighestSalary 
        );     
END

178. 分数排名

# 窗口
select score,
       dense_rank() over(order by Score desc) as 'Rank'
from Scores;
# 清奇思路
select a.Score as Score,
(select count(distinct b.Score) from Scores b where b.Score >= a.Score) as Rank
from Scores a
order by a.Score DESC

179. 最大数

class Solution:
    def largestNumber(self, nums: List[int]) -> str:
        from functools import cmp_to_key
        def helper(x, y):
            if x + y > y + x:
                return -1
            elif x + y < y + x:
                return 1
            else:
                return 0
        return "".join(sorted(map(str, nums), key=cmp_to_key(helper))).lstrip("0") or "0"

180. 连续出现的数字

# pre变量统计上一次Num，cnt统计连续次数
SELECT DISTINCT a.Num ConsecutiveNums FROM (
SELECT t.Num ,
       @cnt:=IF(@pre=t.Num, @cnt+1, 1) cnt,
       @pre:=t.Num pre
  FROM Logs t, (SELECT @pre:=null, @cnt:=0) b) a
  WHERE a.cnt >= 3