Minimum Size Subarray Sum_java minimum size subarray sum-优快云博客

本文链接：https://blog.youkuaiyun.com/guorudi/article/details/45750427

本文探讨了如何在给定数组中找到满足特定条件的最小长度子数组，具体为求和大于等于给定整数的问题。通过双指针法优化算法效率，实现快速查找并返回符合条件的子数组长度。

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.

click to show more practice.

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

双指针，用cmp保存数列的值，每次判断（前一保存数组与当前值的和）与当index前值的大小，如果小于当前index的值，说明前段数组小于零舍弃，从当前index计数。如果cmp大于s，则从left指针开始舍弃，直到cmp小于s，保存数组的长度，具体的代码如下

public class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        if(nums.length<=0||s<=0) return 0;
        int cmp = nums[0], res = 1,minRes = Integer.MAX_VALUE;
        int left = 0;
        if(cmp>=s) return res;
        for(int i=1;i<nums.length;i++){
        	if(nums[i]>=cmp +nums[i]){
        		left = i;
        		cmp = nums[i];
        		res = 1;
        	}else{
        		cmp = cmp+nums[i];
        		res++;
        	}
        	if(cmp>=s){
        		while(cmp>=s){
        			cmp -= nums[left];
        			left++;
        			res--;
        		}
        		minRes = Math.min(res+1, minRes);
        	}
        }
        return minRes==Integer.MAX_VALUE?0:minRes;
    }
}