在合并和排序函数中列出超出范围的索引

Sch*_*ger 1 python mergesort

我尝试在 python 中编写一个简单的合并和排序函数,但在收到以下错误后卡住了 -

List out of range. 
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如果您能帮我修复它并找出如何避免它,我将不胜感激。我在下面添加了代码-

def merge(lst1, lst2):
    # Gets two sorted lists and returns one merged and sorted list
    merge_sorted = []
    i = 0
    j = 0
    len1 = len(lst1) - 1
    len2 = len(lst2) - 1
    while i < len1 or j < len2:
        if lst1[i] < lst2[j]:
            merge_sorted.append(lst1[i])
            i += 1
        elif lst1[i] > lst2[j]:
            merge_sorted.append(lst2[j])
            j += 1
        else:
            merge_sorted.append(lst1[i])
            merge_sorted.append(lst2[j])
            i += 1
            j += 1
    return merge_sorted

lst1 = [2, 4, 5, 6, 8]
lst2 = [1, 3, 7, 9, 0]
merge(lst1, lst2)
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我得到了什么:

IndexError                                Traceback (most recent call last)
<ipython-input-13-572aad47097b> in <module>()
     22 lst1 = [2, 4, 5, 6, 8]
     23 lst2 = [1, 3, 7, 9, 0]
---> 24 merge(lst1, lst2)

<ipython-input-13-572aad47097b> in merge(lst1, lst2)
      7     len2 = len(lst2) - 1
      8     while i < len1 or j < len2:
----> 9         if lst1[i] < lst2[j]:
     10             merge_sorted.append(lst1[i])
     11             i += 1

IndexError: list index out of range
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Pat*_*ner 6

你的问题是 while 条件:

while i < len1 or j < len2:
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它应该是and- 如果其中一个条件不正确,您只需将非空列表的其余部分附加到您的结果中,您就完成了。

您当前的代码仍然进入,而体和检查if lst1[i] < lst2[j]:,如果的一个i/j越大那么你得到你有错误的列表。


完整的固定代码:

def merge(lst1, lst2):
    # Gets two sorted lists and returns one merged and sorted list
    merge_sorted = []
    i = 0
    j = 0
    len1 = len(lst1) - 1
    len2 = len(lst2) - 1
    while i < len1 and j < len2:         # use and
        if lst1[i] < lst2[j]:
            merge_sorted.append(lst1[i])
            i += 1
        elif lst1[i] > lst2[j]:
            merge_sorted.append(lst2[j])
            j += 1
        else:
            merge_sorted.append(lst1[i])
            merge_sorted.append(lst2[j])
            i += 1
            j += 1

    # add remainder lists - the slices evaluate to [] if behind the list lengths
    merge_sorted.extend(lst1[i:])  # if i is aready out of the list this is []
    merge_sorted.extend(lst2[j:]) # if j is aready out of the list this is []
    return merge_sorted

lst1 = [2, 4, 5, 6, 8]
lst2 = [0, 1, 3, 7, 9]  # fixed input, needs to be sorted, yours was not
print(merge(lst1, lst2))
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输出:

[0, 1, 2, 3, 4, 5, 6, 8, 7, 9]
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