Question

Python 2.7 在elif中获取堆栈的索引错误找到字符串中最长的子字符串（仅限非重复字符）返回长度

from pythonds.basic import Stack
def longest_nonrepeating_len(s):
    """find longest non-repeating substring and return its length"""
    if len(s) < 1:
        return []

    longest_substring = 0
    max_long_substring = 0
    stack = Stack()
    stack.push(s[0])

    for char in range(0,len(s)):
        if stack.isEmpty():
            stack.push(s[char])
            longest_substring = 1
            max_long_substring = 1
            continue

        if s[char] != stack.pop():
            longest_substring += 1
            stack.push(s[char])
            max_long_substring = longest_substring
        elif s[char] == stack.pop():
            longest_substring = 0
            stack.push(s[char])

    return max_long_substring

ERROR：

Traceback (most recent call last):
  File "python", line 1, in <module>
  File "python", line 23, in longest_nonrepeating_len
IndexError: pop from empty list

>>> longest_nonrepeating_len("abccd")
3
>>> longest_nonrepeating_len("ffff")
1

Answer 1

当stack只有一个元素时会发生这种情况，因为你连续两次调用stack.pop（）。我想你的意思是：

    pop = stack.pop()
    if s[char] != pop:
        longest_substring += 1
        stack.push(s[char])
        max_long_substring = longest_substring
    elif s[char] == pop:
        longest_substring = 0
        stack.push(s[char])

但是我认为您的代码逻辑仍存在缺陷：在更新之前，您是否应该检查longest_substring实际上是否比之前的max_long_substring值更长？

python - 在字符串中查找最长的非重复substr

1 个答案: