209. Minimum Size Subarray Sum 【M】【35】

寻找最小子数组和问题的双指针解决方案

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本文详细介绍了如何使用双指针技术解决给定数组中找到和大于等于特定值的最小子数组长度的问题。通过遍历并调整两个指针的位置，实现了O(n)的时间复杂度。此外，文章还提供了进一步优化为O(n log n)解决方案的思考过程。该技术适用于数组操作和算法优化领域。

Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn't one, return 0 instead.

For example, given the array [2,3,1,2,4,3] and s = 7,
the subarray [4,3] has the minimal length under the problem constraint.

click to show more practice.

More practice:

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

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

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一个很有趣的双指针

从头到尾遍历，b每次增大一，然后观察a可以减少几个

class Solution(object):
    def minSubArrayLen(self, s, nums):
        if sum(nums) < s:
            return 0

        minn = 2 << 30   
        a,b = 0,0
        total = nums[0]
        res = 0

        while total + nums[b] < s:
            b += 1
            total += nums[b]

        res = b + 1

        while b < len(nums):
            while total - nums[a] >= s:
                total -= nums[a]
                a += 1
            res = b - a + 1
            minn = min(minn,res)
            b += 1
            try:
                total += nums[b]
            except:
                break
        return minn