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Good Job!!! @KL2333Python homework:
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Good Job!!! @KL2333Python homework:
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27感觉BC都可以呀??
Good Job!!! @KL2333Python homework:
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Good Job!!! @KL2333Python homework:
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Bug 已经解决，现在题目总数可以识别为正确的108了哈 |
Good Job!!! @KL2333Python homework:
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错好多.... |
Good Job!!! @KL2333Python homework:
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Good Job!!! @KL2333Python homework:
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Good Job!!! @KL2333Python homework:
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Good Job!!! @KL2333Python homework:
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Good Job!!! @KL2333Python homework:
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Good Job!!! @KL2333Python homework:
Numpy homework:
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Good Job!!! @KL2333Python homework:
Numpy homework:
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Good Job!!! @KL2333Python homework:
Numpy homework:
Pandas homework:
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Good Job!!! @KL2333Python homework:
Numpy homework:
Pandas homework:
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Good Job!!! @KL2333Python homework:
Numpy homework:
Pandas homework:
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Good Job!!! @KL2333Python homework:
Numpy homework:
Pandas homework:
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Good Job!!! @KL2333Python homework:
Numpy homework:
Pandas homework:
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Good Job!!! @KL2333Python homework:
Numpy homework:
Pandas homework:
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Good Job!!! @KL2333Python homework:
Numpy homework:
Pandas homework:
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Good Job!!! @KL2333 |
1 similar comment
Good Job!!! @KL2333 |
[434. Number of Segments in a String](https://leetcode.com/problems/number-of-segments-in-a-string/)class Solution:
def countSegments(self, s: str) -> int:
s=s.strip() # 清理空格
count=0 # 计数器初始化
c_ount=0
if len(s)==0: # 特殊情况
return 0
for char in s: # 循环遍历
if char != ' ': # 空格计数器reset情况
c_ount =0
continue # 同时跳出本次循环
c_ount+=1 # 空格计数器+1
if c_ount==1: # 对连续和普通空格均仅计数一次
count+=1
return count+1 # 空格计数器计数+1为段数 [1869. Longer Contiguous Segments of Ones than Zeros](https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/description/)class Solution:
def checkZeroOnes(self, s: str) -> bool:
a=s.split('0') # 对s按0分段
b=s.split('1') # 对s按1分段
m1=max(len(c) for c in a) # 最长的1的长度
m2=max(len(c) for c in b) # 最长的0的长度
return m1>m2 # 返回两者最长长度的比较[1784. Check if Binary String Has at Most One Segment of Ones](https://leetcode.com/problems/check-if-binary-string-has-at-most-one-segment-of-ones/description/)class Solution:
def checkOnesSegment(self, s: str) -> bool:
if len(s)==1: # 特殊情况
return 1
s=s.rstrip('0') # 清理末尾的'0'
if len(s.split('0'))==1: # 按'0'分割,段数为1即满足要求
return 1
else :
return 0 # 其它类型[852. Peak Index in a Mountain Array](https://leetcode.com/problems/peak-index-in-a-mountain-array/description/)class Solution:
def peakIndexInMountainArray(self, arr: List[int]) -> int:
return arr.index(max(arr)) # 返回峰值的index[162. Find Peak Element](https://leetcode.com/problems/find-peak-element/description/)class Solution:
def findPeakElement(self, nums: List[int]) -> int:
if len(nums)<3: # 特殊情况短的直接判断
return nums.index(max(nums))
nums.insert(0,float('-inf')) # 前后插入负无穷辅助判断两端点
nums.append(float('-inf'))
for i in range(0,len(nums)-1): # 判断相邻三个的大小是否为峰型,由于之前在list前插入了元素,此时的index均要-1
if nums[i]<nums[i+1] and nums[i+1]>nums[i+2]:
return i |
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愚钝如我,暂时放弃 |
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@iphysresearch 可以记录了 |
@KL2333 也有同学遇到类型的情况：#137 (comment) 另外，答案可以互相抄哈。。。只要你找得到哈 |
iphysresearch
modified the milestones:
Leetcode HW finished,
Python/Numpy/Pandas HW completed
Good Job!!! @KL2333Python homework:
Numpy homework:
Pandas homework:
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完成Python扩展作业！ |
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