Leetcode** 209. Minimum Size Subarray Sum

Description: Given an array of positive integers nums and a positive integer target, return the minimal length of a contiguous subarray [numsl, numsl+1, ..., numsr-1, numsr] of which the sum is greater than or equal to target. If there is no such subarray, return 0 instead.

Link: 209. Minimum Size Subarray Sum

Examples:

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

思路: O(n)复杂度，只遍历一遍，不会啊，没思路。这个视频讲解的很好。首先left=0，然后right指针从0向后移，直到sum(nums[l:r+1]) >= target，然后让left从0向右移，缩小这个sum>=target的区间，更新最小区间的值，直到缩小到其和小于target, 然后right指针再开始右移，循环此过程，直到right指针到nums的尾。

class Solution(object):
    def minSubArrayLen(self, target, nums):
        """
        :type target: int
        :type nums: List[int]
        :rtype: int
        """
        l, s = 0, 0
        res = float("inf")
        for i in range(len(nums)):
            s += nums[i]
            while l <= i and s >= target:
                res = min(res, i-l+1)
                s -= nums[l]
                l += 1
        if res == float("inf"):
            return 0
        return res

日期: 2021-04-13