连接 K 周围的字符串的Python程序

给定字符串列表，连接出现在字符串K 周围的所有字符串。

Input : test_list = [“Gfg”, “*”, “is”, “best”, “*”, “love”, “gfg”], K = “*” 
Output : [‘Gfg*is’, ‘best*love’, ‘gfg’] 
Explanation : All elements around * are joined.

Input : test_list = [“Gfg”, “$”, “is”, “best”, “$”, “love”, “gfg”], K = “$” 
Output : [‘Gfg$is’, ‘best$love’, ‘gfg’] 
Explanation : All elements around $ are joined. 
 

方法一：使用循环

这是可以执行此任务的粗暴方式。在这里，我们遍历所有元素并检查 K，如果找到，我们将与前一个和后继元素执行所需的连接。

Python3

# Python3 code to demonstrate working of
# Concatenate Strings on K String
# Using loop
  
# initializing list
test_list = ["Gfg",  "+", "is", "best", "+", "love", "gfg"]
  
# printing original list
print("The original list is : " + str(test_list))
  
# initializing K for Concatenate
K = "+"
  
res = []
idx = 0
  
while idx < len(test_list):
    
    ele = test_list[idx]
  
    # concatenation if next symbol is K
    if (idx < len(test_list) - 1) and test_list[idx + 1] == K:
        ele = ele + K + test_list[idx + 2]
  
        # increasing counter by 2
        idx += 2
    res.append(ele)
    idx += 1
  
# printing result
print("Strings after required concatenation : " + str(res))

输出

The original list is : ['Gfg', '+', 'is', 'best', '+', 'love', 'gfg']
Strings after required concatenation : ['Gfg+is', 'best+love', 'gfg']

方法 2：使用join() + replace() + split()

以上功能的组合可以解决这个问题。在这种情况下，我们执行所有元素的连接，然后删除目标 K 周围的空间。被视为单个字符串，拆分所需的字符串会产生围绕 K 的连接值。

蟒蛇3

# initializing list
test_list = ["Gfg",  "+", "is", "best", "+", "love", "gfg"]
  
# printing original list
print("The original list is : " + str(test_list))
  
# initializing K for Concatenate
K = "+"
  
# performing split after removing space around K
# splits assuming Strings joined around K
res = ' '.join(test_list).replace(' ' + K + ' ', K).split()
  
# printing result
print("Strings after required concatenation : " + str(res))

输出

The original list is : ['Gfg', '+', 'is', 'best', '+', 'love', 'gfg']
Strings after required concatenation : ['Gfg+is', 'best+love', 'gfg']