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给定一位研究者论文被引用次数的数组（被引用次数是非负整数），数组已经按照 升序排列 。编写一个方法，计算出研究者的 h 指数。h 指数的定义: “h 代表“高引用次数”（high citations），一名科研人员的 h 指数是指他（她）的 （N 篇论文中）总共有 h 篇论文分别被引用了至少 h 次。（其余的 N - h 篇论文每篇被引用次数不多于 h 次。）"class Solution: def hIndex(self, citations: List[int]) -> int:

class Solution: def minArray(self, numbers: List[int]) -> int: temp = numbers[0] for i in range(len(numbers)): if temp > numbers[i]: temp = numbers[i] return temp

1、递归2、迭代3、逻辑递归法class Solution: def sumNums(self, n: int) -> int: if n < 1: return 0 return n + self.sumNums(n - 1)s = Solution()print(s.sumNums(5))def sumNums(n): if n == 1: return 1 n += sumNums(n - 1

class Node(): def __init__(self,data): self.data = data self.next = Noneclass stack(): def __init__(self): self.slist = [] def add(self,data): self.slist.append(data) def popdata(self): return self.

#思路：遍历整个链表，把链表中的元素放在列表中，索引第n-k个元素class Node(): def __init__(self,data): self.data = data self.next = Noneclass singleList(): def __init__(self): self._head = None # 尾插法建立列表 def append(self,data): if self._

# 从右上角开始查找def func(target,matrix): rows = len(matrix) cols = len(matrix[0]) x = 0 y = cols - 1 while x < rows and y >= 0: if target == matrix[x][y]: return True elif target < matrix[x][y]: .

使用冒泡排序def find_num(nums,k): for i in range(len(nums)): for j in range(len(nums)-1): if nums[j] >nums[j+1]: nums[j],nums[j+1]=nums[j+1],nums[j] return nums[:k]nums = [2,6,3,1]print(find_num(nums, 2))...

from typing import Listclass Solution: def findRepeatNumber1(self, nums: List[int]) -> int: """ """ dic = set() for val in nums: if val in dic: print(val) dic.add(val) # retur

from typing import Listdef findContinuousSequence(target: int) -> List[List[int]]: i = 1 # 滑动窗口的左边界 j = 1 # 滑动窗口的右边界 sum = 0 # 滑动窗口中数字的和 res = [] while i <= target // 2: if sum < target: # 右边界向右移动

n = int(input("请输入一个整数"))a = input().split(" ")length = [int(a[i]) for i in range(n)]dict1 = {}for i in range(n): dict1[length[i]] = length .count(length[i])a=sorted(dict1.items(),key=lambda item:item[0],reverse= True)print(a)tem=[]for j in

from typing import Listclass Solution: def permuteUnique(self, nums: List[int]) -> List[List[int]]: # 第一次调用i= 0,二次调用i =1 三次调用i = 2 def dfs(nums, size, depth, path, used, res): if depth == size: res.append(path.

题目 合并两个有序列表def merge_list(list1,list2): result = [] while list1 and list2: if list1[0] <= list2[0]: result.append(list1[0]) del list1[0] else: result.append(list2[0]) del list2[0

冒泡排序def bubble_sort(our_list): n = len(our_list) sortBorder = n - 1 # 无序数列边界 for i in range(n): for j in range(0,sortBorder): # 索引 后边有j+1 if our_list[j] > our_list[j+1]: our_list[j],our_list[j+1] =