算法题牛客网NC88 寻找第K大

题目：

思路就是做个排序，要求时间复杂度$O(n\log n)$，因此选用快排。代码：

class Solution:def quickSort(self, a, start, end):if start >= end:returnval = a[start]low = starthigh = endwhile low < high:while low < high and a[high] >= val:high -= 1a[low] = a[high]while low < high and a[low] < val:low += 1a[high] = a[low]a[low] = valself.quickSort(a, start, low-1)self.quickSort(a, low+1, end)def findKth(self , a: List[int], n: int, K: int) -> int:# write code hereself.quickSort(a, 0, n-1)return a[-K]

NC61 两数之和

题目

比较简单，之前也做过，直接用的循环遍历，但本次要求时间复杂度为$O(n\log n)$，循环遍历会超时，参考了下解题思路，用字典做hashmap的方式，代码：

class Solution:def twoSum(self , numbers: List[int], target: int) -> List[int]:# write code here# for i in range(len(numbers)-1):#     for j in range(i+1, len(numbers)):#         if numbers[i] + numbers[j] == target:#             return [i+1, j+1]# return []hash_map = {}for i in range(len(numbers)):tmp = target - numbers[i]if tmp in hash_map:return [hash_map[tmp]+1, i+1]elif numbers[i] not in hash_map:hash_map[numbers[i]] = ireturn []

NC33 合并两个排序的链表

题目：

这个题也是很简单的类型，因为输入就已经排好序了，只要遍历一下链表就可以了。代码：

class Solution:def Merge(self , pHead1: ListNode, pHead2: ListNode) -> ListNode:# write code hereif not pHead1 and not pHead2:return Noneif not pHead1:return pHead2if not pHead2:return pHead1newHead = pHead1 if pHead1.val < pHead2.val else pHead2newCur = newHeadcur1 = pHead1.next if pHead1.val < pHead2.val else pHead1cur2 = pHead2 if pHead1.val < pHead2.val else pHead2.nextwhile cur1 and cur2:if cur1.val < cur2.val:newCur.next = cur1cur1 = cur1.nextelse:newCur.next = cur2cur2 = cur2.nextnewCur = newCur.nextif cur1:newCur.next = cur1if cur2:newCur.next = cur2return newHead

NC76 用两个栈实现队列

题目：

这个题不知道是不是有什么不给用的坑，反正我直接只用了一个栈，过于简单有些怀疑是不是理解有偏差。代码：

class Solution:def __init__(self):self.stack1 = []self.stack2 = []def push(self, node):# write code hereself.stack1.append(node)def pop(self):return self.stack1.pop(0)# return xx