LeetCode 209. Minimum Size Subarray Sum

原题链接在这里:https://leetcode.com/problems/minimum-size-subarray-sum/

题目:

Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn't one, return 0 instead.

For example, given the array [2,3,1,2,4,3] and s = 7,
the subarray [4,3] has the minimal length under the problem constraint.

题解:

类似Minimum Window Substring.

双指针维护一个window, 当window 的 sum小于target时一直移动window 的右边, 更新sum.

当sum >= s后一直移动window 左边, 更新sum 和 minLength.

最后返回时看minLength是否被更新过,若被更新过就返回minLength, 没有更新过说明没有符合要求的window, 返回0.

Time Complexity: O(n), array至多走两遍.

Space: O(1).

AC Java:

 1 class Solution {
 2     public int minSubArrayLen(int s, int[] nums) {
 3         if(nums == null || nums.length == 0 || s <= 0){
 4             return 0;
 5         }
 6         
 7         int walker = 0;
 8         int runner = 0;
 9         int sum = 0;
10         int res = Integer.MAX_VALUE;
11         while(runner < nums.length){
12             sum += nums[runner++];
13             while(sum >= s){
14                 res = Math.min(res, runner-walker);
15                 sum -= nums[walker++];
16             }
17         }
18         
19         return res == Integer.MAX_VALUE ? 0 : res;
20     }
21 }
原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/4868734.html