【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

以你之姓@ 2022-05-22 09:07 149阅读 0赞

# 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