每天(?)一道Leetcode(19) Minimum Size Subarray Sum(2)

Array

209. Minimum Size Subarray Sum

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.

即 求子数组之和大于等于给定值的最小长度
要求实现O(n)和O(logn)两种算法
1.O(n)

2. O(logn)
   利用binary search
   是一种在有序数组中查找某一特定元素的搜索算法。搜索过程从数组的中间元素开始，如果中间元素正好是要查找的元素，则搜索过程结束；如果某一特定元素大于或者小于中间元素，则在数组大于或小于中间元素的那一半中查找，而且跟开始一样从中间元素开始比较。如果在某一步骤数组为空，则代表找不到。这种搜索算法每一次比较都使搜索范围缩小一半

class Solution:
    def minSubArrayLen(self, s, nums):
        """
        :type s: int
        :type nums: List[int]
        :rtype: int
        """
        result = len(nums) + 1
        for idx, n in enumerate(nums[1:], 1):
            nums[idx] = nums[idx - 1] + n
        left = 0
        for right, n in enumerate(nums):
            if n >= s:
                left = self.find_left(left, right, nums, s, n)
                result = min(result, right - left + 1)
        return result if result <= len(nums) else 0

    def find_left(self, left, right, nums, s, n):
        while left < right:
            mid = (left + right) // 2
            if n - nums[mid] >= s:
                left = mid + 1
            else:
                right = mid
        return left