【Python】对于长度为N的数组A，求连续子数组的和最接近0的值

零子数组

对于长度为N的数组A，求连续子数组的和最接近0的值。
如：数组A=[1, -2, 3, 10, -4, 7, 2, -5]它的所有连续子数组中，和最接近0的是哪个？

算法思路

申请比A长1的空间sum[-1,0…,N-1]，sum[i]是A的前i项和。定义sum[-1] = 0
对sum[-1,0…,N-1]排序，然后计算sum相邻元素的差的绝对值，最小值即为所求在A中任意取两个前缀子数组的和求差的最小值。
计算前n项和数组sum和计算sum相邻元素差的时间复杂度，都是O(N)，排序的时间复杂度认为是O(NlogN)，因此，总时间复杂度：O(NlogN)。

Pyhton代码

# 连续子数组的和最接近0的值
def min_subarray(li):
    size = len(li)
    sumli = [0] * (size + 1)
    for i in list(range(size)):
        sumli[i + 1] = sumli[i] + li[i]
    sumli.sort()
    diff = abs(sumli[1] - sumli[0])
    result = diff
    for i in list(range(size)):
        diff = abs(sumli[i + 1] - sumli[i])
        result = min(diff, result)
    return result


# 获取满足条件的连续子数组
def min_subarray2(li):
    size = len(li)
    sumli = [[0, 0]] * (size + 1)  # [0,0] 第二个位置用来存放i
    for i in list(range(size)):
        sumli[i + 1] = [sumli[i][0] + li[i], i + 1]
    print(sumli)
    sumli.sort()
    print(sumli)
    diff = abs(sumli[1][0] - sumli[0][0])  # 初始化
    temp = diff
    j = k = 0
    for i in list(range(size)):
        diff = abs(sumli[i + 1][0] - sumli[i][0])
        if temp > diff:
            temp = diff
            j = sumli[i][1]
            k = sumli[i + 1][1]
    if j > k:
        return li[k:j]
    else:
        return li[j:k]


if __name__ == '__main__':
    li = [1, -2, 3, 10, -4, 7, 2, -5]
    result = min_subarray2(li)
    print(result)
    print(sum(result))

输出结果：

[[0, 0], [1, 1], [-1, 2], [2, 3], [12, 4], [8, 5], [15, 6], [17, 7], [12, 8]]
[[-1, 2], [0, 0], [1, 1], [2, 3], [8, 5], [12, 4], [12, 8], [15, 6], [17, 7]]
[-4, 7, 2, -5]
0