题解 | #未排序数组中累加和小于或等于给定值的最长子数组长度#

import java.util.*;

public class Main{
    public static int maxLen1(int[] arr, int k) {
        if (arr == null || arr.length == 0) {
            return 0;
        }
        int[] minSums = new int[arr.length];
        int[] minSumEnds = new int[arr.length];
        minSums[arr.length - 1] = arr[arr.length - 1];
        minSumEnds[arr.length - 1] = arr.length - 1;
        for (int i = arr.length - 2; i >= 0; i--) {
            if (minSums[i + 1] < 0) {
                minSums[i] = arr[i] + minSums[i + 1];
                minSumEnds[i] = minSumEnds[i + 1];
            } else {
                minSums[i] = arr[i];
                minSumEnds[i] = i;
            }
        }
        int end = 0;
        int sum = 0;
        int res = 0;
        //i是窗口的最左位置,end是窗口最右位置的下一个位置    从左到右遍历
        for (int i = 0; i < arr.length; i++) {
            /*1、如果以i开头的情况下，累加和<=k的最长子数组是arr[i...end-1]，看看是否需要更新res
            2、如果以 i开头的情况下，累加和<=k的最长子数组比arr[i...end-1]，则不会影响结果
            * */
            //以i开始，右扩
            while (end < arr.length && sum + minSums[end] <= k) {
                sum += minSums[end];
                end = minSumEnds[end] + 1;
            }
            res=Math.max(res,end-i);
            if(end>i){
                sum-=arr[i];    //为了下次，以i+1为开头
            }else {     //窗口内部没有数，说明以i开头的所有子数组的累加和都不可能<=k
                end=i+1;
            }
        }

        return res;
    }

    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
        int n=in.nextInt();
        int k=in.nextInt();
        int[] arr=new int[n];
        for (int i = 0; i <n ; i++) {
            arr[i]=in.nextInt();
        }
        int res=maxLen1(arr,k);
        System.out.println(res);
    }
}