滑动窗口算法 - start < n 的条件

Question

下面是一个滑动窗口解决方案，用于从 geeksforgeeks (https://www.geeksforgeeks.org/minimum-length-subarray-sum-greater-given-value/) 中找到总和大于 x 的最小长度子数组

# O(n) solution for finding smallest 
# subarray with sum greater than x 

# Returns length of smallest subarray 
# with sum greater than x. If there  
# is no subarray with given sum, then 
# returns n + 1 
def smallestSubWithSum(arr, n, x): 

    # Initialize current sum and minimum length 
    curr_sum = 0
    min_len = n + 1

    # Initialize starting and ending indexes 
    start = 0
    end = 0
    while (end < n): 

        # Keep adding array elements while current 
        # sum is smaller than x 
        while (curr_sum <= x and end < n): 
            curr_sum += arr[end] 
            end+= 1

        # If current sum becomes greater than x. 
        while (curr_sum > x and start < n): 

            # Update minimum length if needed 
            if (end - start < min_len): 
                min_len = end - start  

            # remove starting elements 
            curr_sum -= arr[start] 
            start+= 1

    return min_len  

我已经测试过这个解决方案可以工作，但我很困惑为什么在最后一个 while 循环中，检查 start 是否小于 n - 你不希望它小于 end，否则 start 可以去超越终点，这对我来说真的没有意义吗？

Answer 1

由于 curr_sum 是通过将元素添加到 end 来构建的，因此它将在 start 到达 end 之前达到零（或小于 x）。这将退出 while 循环。这也意味着该算法可能不适用于数组中的负数。

就我个人而言，我会写的有点不同。这是一个考虑了负数条件的示例：

def minSub(arr,x):
    subTotal      = 0
    size,minSize  = 0,len(arr)+1
    start         = iter(arr)
    for value in arr:
        subTotal += value
        size     += 1
        while subTotal not in range(0,x+1):
            if subTotal>x :
                minSize = min(minSize,size)
            subTotal -= next(start,0)
            size     -= 1
    return minSize

输出：

arr  = [1, 4, 45, 6, 0, 19] 
x    = 51 
print(minSub(arr,x)) #3

arr = [-8, 1, 4, -1, 3, -6]
x   = 6 
print(minSub(arr,x)) # 4

arr = [1, 11, 100, 1, 0, 200, 3, 2, 1, 250]
x   = 280 
print(minSub(arr,x)) # 4

arr = [1, 10, 5, 2, 7]
x   = 9
print(minSub(arr,x)) # 1

arr = [1, 2, 4]
x   = 8
print(minSub(arr,x)) # 4