CareerCup Given an array having positive integers, find a subarray which adds to a given number

最新推荐文章于 2022-04-19 00:41:43 发布
转载 最新推荐文章于 2022-04-19 00:41:43 发布 · 747 阅读 · 0

本文介绍了一种在正整数数组中找到连续子数组使其和等于给定数值的方法。该算法采用一次遍历的方式，通过调整运行总和来确保效率达到O(n)级别。

Given an array having positive integers, find a continous subarray which adds to a given number.

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Take notice that all numbers are positive!

It can be done O(n). 
My logic is as follows:- Keep adding elements to a running sum till it is either equal to given number or is greater than the given number. As soon as we add a number and the running number is greater than the given number we subtract the elements from the beginning of the array till it is again less than the given number. After the current sum is less than the given number we add the next element and check and continue the same process.

Matlab：成功解决Index must be a positive integer or logical
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This problem can be solved using the sliding window technique. We maintain a window of elements whose sum is greater than or equal to the target. We start with two pointers, left and right, both pointing to the first element of the array. We then move the right pointer to the right until the sum of elements in the window is greater than or equal to the target. Once we have found such a window, we update the answer with the length of the current window. We then move the left pointer to the right until the sum of elements in the window is less than the target. We repeat this process until the right pointer reaches the end of the array. We return the answer. Here is the Python code: ``` def minSubArrayLen(nums, target): n = len(nums) left = right = 0 window_sum = 0 ans = float('inf') while right < n: window_sum += nums[right] while window_sum >= target: ans = min(ans, right - left + 1) window_sum -= nums[left] left += 1 right += 1 return ans if ans != float('inf') else 0 ``` The time complexity of this algorithm is O(n) and the space complexity is O(1).
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