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作者: 负雪明烛
id: fuxuemingzhu
个人博客: http://fuxuemingzhu.cn/
题目地址: https://leetcode.com/problems/minimum-size-subarray-sum/description/
题目描述:
Given an array of n
positive integers and a positive integer s, find the minimal length of a contiguous
subarray of which the sum ≥ s
. If there isn’t one, return 0 instead.
Example:
Input: s = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: the subarray [4,3] has the minimal length under the problem constraint.
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).
题目大意
找出一个数组中最短连续的子数组,这个子数组的和要>=s.
解题方法
碰巧今天在《挑战程序设计竞赛》一书中看到这个题,解法称之为虫取法,其实就是双指针。其实看到让连续子数组满足一定条件的很多都用了双指针,比如713. Subarray Product Less Than K。
因为这个题需要求最小值,所以结果初始化为inf,每次移动一下右指针,当和满足条件的时候,更新结果,并移动左指针,同时记得把和删去左边的数字。
时间复杂度是O(N),空间复杂度是O(1)。
class Solution:
def minSubArrayLen(self, s, nums):
"""
:type s: int
:type nums: List[int]
:rtype: int
"""
N = len(nums)
l, r = 0, 0
csum = 0
res = float('inf')
while r < N:
csum += nums[r]
while csum >= s:
res = min(res, r - l + 1)
csum -= nums[l]
l += 1
r += 1
return res if res != float('inf') else 0
参考资料:
日期
2018 年 10 月 15 日 —— 美好的周一怎么会出现雾霾呢?