Leetcode/ and continuous shortest subarray greater than or equal to target

Code

package com.xcrj;

/** *  The finger of the sword  Offer II 008.  And greater than or equal to target Continuous shortest subarray of  *  Given a containing n An array of positive integers and a positive integer  target . *  Find the sum of the array  ≥ target  The smallest length of   Continuous subarray [numsl, numsl+1, ..., numsr-1, numsr] , And return its length . If there is no sub array that meets the conditions , return  0 . *  Elements and >=target&& The subsequence of elements is continuous $$ The length of subsequence is the shortest  */
public class Solution8 {
    
    /** *  The sliding window ( Double pointer ) * ！！！ Use different pointers to achieve different purposes  * i j It all points to 0 *  Move first j  Seek sum>=target *  Move again i  Seek sum<target *  Under the resultant force of the two phases, the subsequence with the smallest length satisfying the condition is obtained  */
    public int minSubArrayLen(int target, int[] nums) {
    
        int i = 0;
        int j = 0;
        int sum = 0;
        int minLen = Integer.MAX_VALUE;
        //
        while (j < nums.length) {
    
            sum += nums[j];
            while (sum >= target) {
    
                minLen = Math.min(minLen, j - i + 1);
                sum -= nums[i];
                //  Move i  Seek sum<target
                i++;
            }
            //  Move j  Seek sum>=target
            j++;
        }

        return minLen == Integer.MAX_VALUE ? 0 : minLen;
    }

    public static void main(String[] args) {
    
        Solution8 solution8 = new Solution8();
        System.out.println(solution8.minSubArrayLen(7, new int[]{
    2, 3, 1, 2, 4, 3}));
    }
}

Reference resources

author ：LeetCode-Solution
link ：https://leetcode.cn/problems/2VG8Kg/solution/he-da-yu-deng-yu-target-de-zui-duan-zi-s-ixef/
source ： Power button （LeetCode）