合并排序列表索引超出范围

Question

I'm trying to do a basic merge sort in python, but every time I run the code I get the error

"... if x[0] < y[0]: IndexError: list index out of range "

And I'm not sure where I'm going wrong. Here's the code:

def merge(x,y):
    merged = []
    while len(x) > 0 or len(y) > 0:
        if x[0] < y[0]: #this is where it is telling me the list index is out of range
            merged.append(x[0])
            del x[0] 
        else:
            merged.append(y[0])
            del y[0]
    print merged

s1 = raw_input()
s2 = raw_input()
nums1 = map(int, s1.split())
nums2 = map(int, s2.split())

merge(nums1,nums2)

Thank you in advance!

Answer 1

Once you've taken all the elements from x , there is no x[0] , but you try to compare it to y[0] anyway. If you exhaust y first, there is no y[0] , but you still try to compare it to x[0] anyway. Once one list is exhausted, you can't keep doing those comparisons.

Answer 2

Your current while statement will run while either of the two lists are non-empty (because of the or condition) but you're not covering the case where just one is empty. In the case where one list is empty, and you attempt to compare the first in each list, you'll get the error you're seeing.

For example, consider the simple lists {1,2} and {3,4} . You start with:

list 1 = {1,2}, list 2 = {3,4}, merged = {}

Your first loop iteration will check the first element of both lists and decide it needs to use the first list, leaving you with:

list 1 = {2}, list 2 = {3,4}, merged = {1}

Your second loop iteration will check the first element of both lists and decide it needs to use the first list again, leaving you with:

list 1 = {}, list 2 = {3,4}, merged = {1,2}

Your third loop iteration will check the first element of both lists and fall in a heap because there is no first element in the first list.

The solution is to simply ensure that, if one list has become empty, just use the other list without checking the contents. There are two ways I've done that in the past.

First, within the while loop, you can check if each list is empty before comparing. If one is, select from the other list. In other words, pseudocode like:

while len(list1) > 0  or len(list2) > 0:
    if    len(list1) == 0:       use list2
    elsif len(list2) == 0:       use list1
    elsif list1[0] <= list2[0]:  use list1
    else:                        use list2

Second, and this is the one I tend to prefer since it seems cleaner having the finishing logic outside of the main loop, compare lists until one is empty, then process the rest of the other list:

while len(list1) > 0  and len(list2) > 0:
    if    list1[0] <= list2[0]:  use list1
    else:                        use list2

 while len(list1) > 0:  use list1
 while len(list2) > 0:  use list2

Only one of those final while loops will do anything since you only get to that point if one of the lists is empty.

You can also optimise it in languages that support mass transfer since you know you have to transfer a large block from one array to another (you don't need to do it one element at a time). I'm referring there to something like Python's:

dst[28:31] = list2[12:15]

or C's memcpy() function.

The last section then becomes something like:

 if len(list1) > 0:  use rest of list1
 else:               use rest of list2

It's a simple "either-or" since you know:

• at least one of the lists is empty, otherwise you'd still be in the main loop; and
• you know the other list is non-empty since you're only pulling an element from one list each time.