Homework

[Pdirac52]

try python, numpy and pandas homework

[iphysresearch]

Good Job!!! @Pdirac52

[iphysresearch]

Good Job!!! @Pdirac52

@Pdirac52
你需要update branch 一下哈！ #137 (comment)

[iphysresearch]

Good Job!!! @Pdirac52

Python homework:

• Total questions: 108
• Correct answers: 1
• Score: 0%

---

Numpy homework:

• Total questions: 10
• Correct answers: 10
• Score: 100.00%

---

Pandas homework:

• Total questions: 12
• Correct answers: 12
• Score: 100.00%

---

[iphysresearch]

Good Job!!! @Pdirac52

Python homework:

• Total questions: 108
• Correct answers: 0
• Score: 0%

---

Numpy homework:

• Total questions: 10
• Correct answers: 10
• Score: 100.00%

---

Pandas homework:

• Total questions: 12
• Correct answers: 12
• Score: 100.00%

---

[iphysresearch]

Good Job!!! @Pdirac52

Python homework:

• Total questions: 108
• Correct answers: 108
• Score: 100.00%

---

Numpy homework:

• Total questions: 10
• Correct answers: 10
• Score: 100.00%

---

Pandas homework:

• Total questions: 12
• Correct answers: 12
• Score: 100.00%

---

[iphysresearch]

Good Job!!! @Pdirac52

Python homework:

• Total questions: 108
• Correct answers: 108
• Score: 100.00%

---

Numpy homework:

• Total questions: 10
• Correct answers: 10
• Score: 100.00%

---

Pandas homework:

• Total questions: 12
• Correct answers: 12
• Score: 100.00%

---

[Pdirac52]

王老师，我想提交一下python，numpy，pandas基础作业

[iphysresearch]

@Pdirac52
很好哈！把Python编程的扩展作业也做一下吧哈？

[iphysresearch]

Good Job!!! @Pdirac52

Python homework:

• Total questions: 108
• Correct answers: 108
• Score: 100.00%

---

Numpy homework:

• Total questions: 10
• Correct answers: 10
• Score: 100.00%

---

Pandas homework:

• Total questions: 12
• Correct answers: 12
• Score: 100.00%

---

[Pdirac52]

王老师好，是这样在comment中提交leetcode作业对么

Leetcode:

No.434:

class Solution:
    def countSegments(self, s):
        segment_count = 0     #先定义初始计数为0

        for i in range(len(s)):    #做循环，遍历该字符串每一个字符
            if (i == 0 or s[i - 1] == ' ') and s[i] != ' ':#当i-1是空格，且i不是空格时，认为i是一个单词的开头
                segment_count += 1   #因此单词数加一

        return segment_count

No.1869

class Solution:
    def checkZeroOnes(self, s: str) -> bool:
        # 初始化变量
        max0=0           
        max1=0
        length0=0
        length1=0     
        # 遍历字符串
        for i in s:
            if i=='1':       #判断此时是不是1
                length1 +=1   #如果是则给加上1，就能数出目前的1子串有多长
                max0=max(length0,max0) #同时给上一个0的最长子串存档并且清零临时子串
                length0=0
            elif i=='0':    #对于0的做法同理
                length0 +=1
                max1=max(length1,max1)
                length1=0
        max1=max(length1,max1) #最后得又赋值一遍，因为如果子串全是1，则不会进入elif的0，则没有给1存入最长子串
        max0=max(length0,max0) #所以需要又存一次
        return max1>max0

No.1784

class Solution:
    def checkOnesSegment(self, s: str) -> bool:
        return '01' not in s #只需要保证01不在字符串中就行

No.852

class Solution:
    def peakIndexInMountainArray(self, arr: List[int]) -> int:
        #题目提示已经保证了是一个山脉数组。即我找到后一个小于前一个的地方为山顶
        for i in range(len(arr)):
            if arr[i]>arr[i+1]:
                return i
#但是感觉还可以更快，用二分法，不断的取中间值来判断应该会更快

No.162

class Solution:
    def findPeakElement(self, nums: List[int]) -> int:
        #思路，不断的取中间值来寻找
        #先找到中间值，然后判断是否是峰值
        #如果是则已经找到
        #如果不是，则看中间值两边谁大，然后往大的那边去靠近
        #因为题目给的列表一定是两边小的
        n=len(nums) 
        def get(i): #方便处理边界，我们让边界是负无穷，不影响数组整体的山峰
            if i==-1 or i==n:
                return float('-inf')
            return nums[i]
            #初始化我们左右取值
        left,right,ans =0,n-1,-1
        while left<=right: 
                #保证左边小于等于右边      
            mid =(left+right)//2
                #然后去判断中间值是否是山峰
            if get(mid-1)<get(mid)>get(mid+1):  
                ans=mid
                break
                    #如果不是山峰，则左右往是山峰的那一次靠近
            if get(mid)<get(mid+1):   
                left=mid+1
            else:
                right=mid-1
        return ans

[iphysresearch]

@Pdirac52
恭喜完成Python编程的基础作业和扩展作业！