LeetCode 209 Minimum Size Subarray Sum 题目解析和python代码

题目：
Given an array of positive integers nums and a positive integer target, return the minimal length of a
subarray
whose sum is greater than or equal to target. If there is no such subarray, return 0 instead.

Example 1:

Input: target = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: The subarray [4,3] has the minimal length under the problem constraint.
Example 2:

Input: target = 4, nums = [1,4,4]
Output: 1
Example 3:

Input: target = 11, nums = [1,1,1,1,1,1,1,1]
Output: 0

Constraints:

1 <= target <= 109
1 <= nums.length <= 105
1 <= nums[i] <= 104

Follow up: If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log(n)).

题目解析：
这里我们可以使用 sliding window 的技巧。
我们可以使用两个 pointers，一个是 start 一个是 end，两个指针都从array的开头开始。
向右移动 end 指针来提高 window 的大小，每一次移动指针都把 nums[end] 添加到现在的和。
当这个和大于或等于 target 时，我们要开始缩短 subarray 的长度。于是我们开始移动 start 这个指针来缩小 window 的大小。通过不停的向右移动 start 指针，直到找到最短的 subarray 的长度。

class Solution:
    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
        start = 0
        curr_sum = 0
        min_len = float('inf')

        for end in range(len(nums)):
            curr_sum += nums[end]

            while curr_sum >= target:
                min_len = min(min_len, end - start + 1)
                curr_sum -= nums[start]
                start += 1
            
        return min_len if min_len != float('inf') else 0

Time complexity 是 O(n)。
Space complexity 是 O(1)。