[leetcode] 209. Minimum Size Subarray Sum @ python

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原题

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).

解法1

双指针法. 用l, r代表左右两个指针, 使用while循环, 累加total, 当total >= s时, 尽量把l往左移, 找出符合题意的最短的subarray, 跳出内层循环后右指针继续往右走, 直到刷完nums.
Time: O(n)
Space: O(1)

代码

class Solution:
    def minSubArrayLen(self, s, nums):
        """
        :type s: int
        :type nums: List[int]
        :rtype: int
        """
        if sum(nums) < s: return 0
        res = len(nums)+1
        l, r, total = 0, 0, 0
        while r < len(nums):
            total += nums[r]
            while total >= s:
                # move left to the most-right
                res = min(res, r-l+1)
                total -= nums[l]
                l += 1
            r += 1
        return res if res != len(nums)+1 else 0

解法2

思路同解法1, 只不过用enumerate()函数简化代码

代码

class Solution:
    def minSubArrayLen(self, s, nums):
        """
        :type s: int
        :type nums: List[int]
        :rtype: int
        """
        if sum(nums) < s: return 0
        res = len(nums)
        left, total = 0, 0
        for i, n in enumerate(nums):
            total += n
            while total >= s:
                res = min(res, i-left +1)
                total -= nums[left]
                left += 1
        return res