python - 查找包含集合中所有值的最短连续子数组的算法

Question

我有以下问题需要解决:

给定一组整数，例如{1,3,2}，以及一个随机整数数组，例如

[1, 2, 2, -5, -4, 0, 1, 1, 2, 2, 0, 3,3]

找到包含集合中所有值的最短连续子数组。如果找不到子数组，则返回一个空数组。

结果:[1, 2, 2, 0, 3]

或者

[1, 2, 2, -5, -4, 3, 1, 1, 2, 0], {1,3,2}。

结果:[3, 1, 1, 2]

我已经尝试了以下设置，我的第二个循环似乎有问题。我不确定我需要更改什么:

def find_sub(l, s):
    i = 0
    counts = dict()
    end = 0
    while i < len(s):
        curr = l[end]
        if curr in s:
            if curr in counts:
                counts[curr] = counts[curr] + 1
            else:
                counts[curr] = 1
                i += 1
        end += 1
    curr_len = end

    start = 0
    for curr in l:
        if curr in counts:
            if counts[curr] == 1:
                if end < len(l):
                    next_item = l[end]
                    if next_item in counts:
                        counts[next_item] += 1
                    end += 1
            else:
                counts[curr] -= 1
                start += 1
        else:
            start += 1
    if (end - start) < curr_len:
        return l[start:end]
    else:
        return l[:curr_len]

Answer 1

您正在使用双指针方法，但只移动两个索引一次 - 直到找到第一个匹配项。您应该重复 move right - move left 模式以获得最佳索引间隔。

def find_sub(l, s):
    left = 0
    right = 0
    ac = 0
    lens = len(s)
    map = dict(zip(s, [0]*lens))
    minlen = 100000
    while left < len(l):
        while right < len(l):
            curr = l[right]
            right += 1
            if curr in s:
                c = map[curr]
                map[curr] = c + 1
                if c==0:
                    ac+=1
                    if ac == lens:
                        break
        if ac < lens:
            break

        while left < right:
            curr = l[left]
            left += 1
            if curr in s:
                c = map[curr]
                map[curr] = c - 1
                if c==1:
                    ac-=1
                    break

        if right - left + 1 < minlen:
            minlen = right - left + 1
            bestleft = left - 1
            bestright = right

    return l[bestleft:bestright]

print(find_sub([1, 2, 2, -5, -4, 3, 1, 0, 1, 2, 2, 0, 3, 3], {1,3,2}))
print(find_sub([1, 2, 2, -5, -4, 3, 1, 0, 1, 2, 2, 1, 0, 3, 3], {1,3,2}))
>>[2, -5, -4, 3, 1]
>>[2, 1, 0, 3]