刷题记录帖子

Leetcode 2140. Solving Questions With Brainpower

感觉中文力扣的解答真的不错，都是直击心灵的。

推荐这个大神，真是大神，给出的解答都是直接心灵的。

比如这道题目，逆向 DP，

正向 DP

1041. Robot Bounded In Circle

这个题目其实不难，但是难度在于如何方便快速的判断和模拟。

主要点事两个，把机器人的属性拆成两个，机器人脸的朝向 和 位置(x,y)

1. class Solution:
   
2.     def isRobotBounded(self, instructions: str) -> bool:
   
3.         dirs = [(0,1),(-1,0),(0,-1),(1,0)]
   
4.         d = 0
   
5.         x = y = 0
   
6.         for op in instructions:
   
7.             if op == "G":
   
8.                 x += dirs[d][0]
   
9.                 y += dirs[d][1]
   
10.             elif op == "L":
    
11.                 d = (d + 1) % 4
    
12.             else:
    
13.                 d = (d - 1) % 4
    
14.             print(x,y,d)
    
15.         return x == y == 0 or d != 0

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2359. Find Closest Node to Given Two Nodes

周赛题目，比较坑爹，这个题目的理解问题。

such that the maximum between the distance from node1 to that node, and from node2 to that node is minimized.

这个其实是，

min of max( value1, value2)

1472. Design Browser History

1. class BrowserHistory:
   
2. 
   
3.     def __init__(self, homepage: str):
   
4.         self.history_stack = [homepage]
   
5.         self.current = 0        
   
6. 
   
7.     def visit(self, url: str) -> None:
   
8.         if self.current >= 0:
   
9.             self.history_stack = self.history_stack[:self.current+1]
   
10.         self.history_stack.append(url)
    
11.         self.current += 1
    
12. 
    
13.     def back(self, steps: int) -> str:
    
14.         if steps <= self.current:
    
15.             self.current = self.current - steps
    
16.         else:
    
17.             self.current  = 0
    
18.         #print(self.history_stack)  
    
19.         #print(self.current)         
    
20.         return self.history_stack[self.current]
    
21.         
    
22.     def forward(self, steps: int) -> str:
    
23.         if steps <= (len(self.history_stack) - 1 - self.current):
    
24.             self.current = self.current + steps
    
25.         else:
    
26.             self.current  = len(self.history_stack) - 1
    
27.         return self.history_stack[self.current]
    
28.         
    
29. 
    
30. 
    
31. # Your BrowserHistory object will be instantiated and called as such:
    
32. # obj = BrowserHistory(homepage)
    
33. # obj.visit(url)
    
34. # param_2 = obj.back(steps)
    
35. # param_3 = obj.forward(steps)

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1. # LC 3417 Zigzag Grid Traversal With Skip<div class="blockcode"><blockquote># LC 3417 Zigzag Grid Traversal With Skip
   
2. 
   
3. class Solution:
   
4.     def zigzagTraversal(self, grid: List[List[int]]) -> List[int]:
   
5.         curr_x, curr_y = 0, 0
   
6.         len_y, len_x = len(grid), len(grid[0])
   
7.         res = list()
   
8.         direction = 1
   
9. 
   
10.         while curr_x < len_x and curr_y < len_y:
    
11.             res.append(grid[curr_y][curr_x])
    
12.             curr_x = curr_x + direction * 2
    
13.             while curr_x < len_x and curr_x >= 0:
    
14.                 res.append(grid[curr_y][curr_x])
    
15.                 curr_x = curr_x + direction * 2
    
16.             curr_y += 1
    
17.             direction *= -1
    
18.             if curr_x == -2: curr_x = 1
    
19.             if curr_x == len_x + 1: curr_x = len_x - 2
    
20.             if curr_x == -1: curr_x = 0
    
21.             if curr_x == len_x: curr_x = len_x -1
    
22.             
    
23. 
    
24.         return res
    
25. 
    
26. 
    
27.             
    
28. # A more elegent solution
    
29. # using i % 2 and the iteration to control order and reversed order of list
    
30. 
    
31. class Solution:
    
32.     def zigzagTraversal(self, grid: List[List[int]]) -> List[int]:
    
33.         res = []
    
34.         for i, r in enumerate(grid):
    
35.             res += r[::-1] if i % 2 else r
    
36.         return res[::2]

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1. # 3263. Convert Doubly Linked List to Array I
   
2. 
   
3. """
   
4. # Definition for a Node.
   
5. class Node:
   
6.     def __init__(self, val, prev=None, next=None):
   
7.         self.val = val
   
8.         self.prev = prev
   
9.         self.next = next
   
10. """
    
11. class Solution:
    
12.     def toArray(self, root: 'Optional[Node]') -> List[int]:
    
13.         node = root
    
14.         res = []        
    
15.         while node:
    
16.             res.append(node.val)
    
17.             node = node.next
    
18.         return res
    
19.         

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1. # 2879. Display the First Three Rows
   
2. 
   
3. import pandas as pd
   
4. 
   
5. def selectFirstRows(employees: pd.DataFrame) -> pd.DataFrame:
   
6.     return employees.head(3)

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LC 1265

1. # """
   
2. # This is the ImmutableListNode's API interface.
   
3. # You should not implement it, or speculate about its implementation.
   
4. # """
   
5. # class ImmutableListNode:
   
6. #     def printValue(self) -> None: # print the value of this node.
   
7. #     def getNext(self) -> 'ImmutableListNode': # return the next node.
   
8. 
   
9. class Solution:
   
10.     def printLinkedListInReverse(self, head: 'ImmutableListNode') -> None:
    
11.         # res = []
    
12.         # curr = head
    
13.         # while curr:
    
14.         #     res.append(curr.printValue())
    
15.         #     curr = curr.getNext()
    
16.         # print(res)
    
17.         # return [res[x] for x in range(len(res)-1, -1, -1)]
    
18. 
    
19.         '''
    
20.         # Recursion
    
21.         if head is not None:
    
22.             self.printLinkedListInReverse(head.getNext())
    
23.             head.printValue()
    
24.         '''
    
25. 
    
26.         # stack
    
27.         stack = []
    
28. 
    
29.         while head:
    
30.             stack.append(head)
    
31.             head = head.getNext()
    
32.         
    
33.         while stack:
    
34.             node = stack.pop()
    
35.             node.printValue()

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1. class Solution:
   
2.     def getSneakyNumbers(self, nums: List[int]) -> List[int]:
   
3.         ss = set()
   
4.         res = []
   
5.         for n in nums:
   
6.             if n not in ss:
   
7.                 ss.add(n)
   
8.             else:
   
9.                 res.append(n)
   
10.         return res

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LC 3289

1. class Solution:
   
2.     def countValidSelections(self, nums: List[int]) -> int:
   
3.         res = 0
   
4.         l, r = 0, sum(nums)
   
5.         for i in range(len(nums)):
   
6.             if nums[i] == 0:
   
7.                 if l == r : res += 2
   
8.                 if l - 1 == r or l == r-1: res += 1
   
9.             else:
   
10.                 l += nums[i]
    
11.                 r -= nums[i]
    
12.         return res

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LC 3354

sum_left --> sum of all elements in left to nums[i]
sum_right --> sum of all elements in right to nums[i]
when nums[i] ==0 and sum_right == sum_left : count+=2
when nums[i] ==0 and sum_right-1==sum_left : count +=1
when nums[i] == 0 and sum_right == sum_left-1 : count +=1