[LeetCode] 209. Minimum Size Subarray Sum (Easy version LC 862)

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于 2018-11-26 09:39:36 发布

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原文链接：https://segmentfault.com/a/1190000017139865

本文介绍了一种在给定正整数数组中寻找最小子数组的方法，该子数组的和大于或等于目标值。通过使用滑动窗口技术，实现了一个时间复杂度为O(n)的解决方案，并探讨了更高效的O(n log n)算法的可能性。

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Problem

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).

Solution

class Solution {
    public int minSubArrayLen(int target, int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        int i = 0, j = 0, sum = 0, min = Integer.MAX_VALUE;
        
        while (i <= j && j < nums.length) {
            sum += nums[j];
            while (sum >= target && i <= j) {
                min = Math.min(min, j-i+1);
                sum -= nums[i];
                i++;
            }
            j++;
        }
        
        return min == Integer.MAX_VALUE ? 0 : min;
    }
}