1) 在字串之間加入空格
txt = "TodayIsFriday" #Today is Friday first = True for letter in txt: if letter>='A' and letter<='Z': if first==True: first = False else: print(' ',end='') print(letter,end='')
輸出:
今天是星期五
2) 刪除字串之間的空格
txt = " Today Is Friday" #Today is Friday for letter in txt: if letter==' ': pass else: print(letter,end='')
輸出:
今天是星期五
3) ltrim- 刪除字串左側的空格。
#ltrim txt = " Today Is Friday" #Today is Friday alphabet = False for letter in txt: if letter==' ' and alphabet == False: pass else: alphabet = True print(letter,end='')
4) rtrim- 刪除字串右側的空格。
txt = "Today Is Friday " #Today is Friday alphabet = False i = len(txt)-1 while i>=0: letter = txt[i] if letter==' ' and alphabet == False: pass else: alphabet = True end = i j = 0 while j<=end: print(txt[j],end='') j+=1 break i-=1
輸出:
今天是星期五
5) 從給定字串中刪除不需要的空格
txt = "Today Is Friday" #Today is Friday i = 0 while i<len(txt): if txt[i] != ' ': print(txt[i],end='') else: if txt[i-1]!=' ': print(txt[i],end='') i+=1
輸出:
今天是星期五
遞迴:
函數呼叫自身。
循環-->迭代方法。
遞歸-->遞歸方法。
範例:1
def display(no): print(no, end=' ') no+=1 if no<=5: display(no) display(1)
輸出:
1 2 3 4 5
呼叫階乘的遞歸函數:
5! =5x4x3x2x1(或)5x4!
4! =4x3x2x1(或)4x3!
3! =3x2x1(或)3x2!
2! =2x1(或)2x1!
1! =1
範例:2
def find_fact(no): if no==1: return 1 return no * find_fact(no-1) result = find_fact(5) print(result)
輸出:
120
說明:
1) find_fact(5)
回傳 5 * find_fact(4) #no-1 = 5-1 -->4
2) find_fact(4)
回傳 4 * find_fact(3) #no-1 = 4-1 -->3
3) find_fact(3)
回傳 3 * find_fact(2) #no-1 = 3-1 -->2
4) find_fact(2)
回傳 2 * find_fact(1) #no-1 = 2-1 -->1
5) find_fact(1)
基本情況:回傳 1
基本情況:遞迴中的基本情況是停止遞迴呼叫的條件。
任務:
strip() - 刪除字串開頭和結尾的所有空白字元(空格、製表符、換行符)。
1) 刪除給定字串前後不需要的空格。
txt = " Today Is Friday " start = 0 end = len(txt) - 1 while start < len(txt) and end >= 0: i = start while i < len(txt) and txt[i] == ' ': i += 1 start = i j = end while j >= 0 and txt[j] == ' ': j -= 1 end = j break result = txt[start:end+1] print(result)
輸出:
Today Is Friday
2)使用遞歸函數反轉數字:
def reverse_a_no(no,reverse = 0): if no==0: return reverse rem = no%10 reverse = (reverse*10) + rem no=no//10 return reverse_a_no(no,reverse) no = int(input("Enter no. ")) reversed_no = reverse_a_no(no) print(reversed_no)
輸出:
Enter no. 15 51
3)是否找到質數:
def find_prime(no,div=2): if div<no: if no%div == 0: return False div+=1 return find_prime(no,div) else: return True no=int(input("Enter the number: ")) if find_prime(no): print("EMIRP number") else: print("not EMIRP number")
輸出:
1) Enter the number: 11 EMIRP number 2) Enter the number: 15 not EMIRP number
4) 找出斐波那契數:
def find_fibonacci(first_num,sec_num,no): if first_num > no: return print(first_num, end=" ") find_fibonacci(sec_num,first_num+sec_num,no) no = int(input("Enter the number: ")) find_fibonacci(0,1,no)
輸出:
Enter the number: 10 0 1 1 2 3 5 8
5。是否查找回文:
def palindrome(num,count=0): if num == 0: return count return palindrome(num//10,count*10+num%10) num=int(input("Enter the number:")) result=palindrome(num) if result==num: print("Palindrome") else: print("Not palindrome")
輸出:
Enter the number:121 Palindrome
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