Python - 重新排列元素第二个索引大于第一个

给定 2 个列表，对于给定的索引，第 2 个列表元素总是大于第一个，如果不是，我们重新排列它。

Input : test_list1 = [36, 38, 40, 132], test_list2 = [35, 37, 39, 41, 133]
Output : [37, 39, 41, 133]
Explanation : Each element in result list is greater than its index counterpart of 1st list. (Eg. 37 > 36)

Input : test_list1 = [2, 6], test_list2 = [5, 3, 8]
Output : [5, 8]
Explanation : Here 5 > 2 and 8 > 6.

方法：使用循环
这是解决这个问题的粗暴方法。在此，我们尝试在整个列表遍历之后获得最合适的下一个更高元素并执行必要的重新排列。

# Python3 code to demonstrate working of 
# Rearrange elements second index greater than first
# Using loop
  
# initializing lists
test_list1 = [14, 16, 18, 110]
test_list2 = [13, 15, 17, 19, 111]
  
  
# printing original lists
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))
  
# Rearrange elements second index greater than first
# Using loop
x = y = 0
res1, res2 = [], []
while x < len(test_list2) and y < len(test_list1):
      
    # checking for greater element
    if test_list2[x] > test_list1[y]:
        res2.append(test_list2[x])
        res1.append(test_list1[y])
        while y < len(test_list1) and test_list2[x] > test_list1[y]:
            res1[-1] = test_list1[y]
            y += 1
    x += 1
  
# printing result 
print("List 2 after conversion : " + str(res2))

输出 ：

The original list 1 is : [14, 16, 18, 110]
The original list 2 is : [13, 15, 17, 19, 111]
List 2 after conversion : [15, 17, 19, 111]