795. Number of Subarrays with Bounded Maximum

We are given an array A of positive integers, and two positive integers L and R (L <= R).

Return the number of (contiguous, non-empty) subarrays such that the value of the maximum array element in that subarray is at least L and at most R.

Example :

Input:
A = [2, 1, 4, 3]
L = 2
R = 3
Output: 3

Explanation:

There are three subarrays that meet the requirements: [2], [2, 1], [3].

Note:

L, R and A[i] will be an integer in the range [0, 10^9].
The length of A will be in the range of [1, 50000].

  遍歷陣列，將遍歷的元素分為下列三種情況來討論：
①：if(A[i] >= L && A[i] <= R)

  在這種情況下，子陣列的種類res += i - j + 1。即每新增一個滿足①條件的點子列的情況就增加 i - j + 1種（j為當前子陣列的開頭，i為當前子陣列的結尾，注意A[j]嚴格滿足A[j] >= L && A[j] <= R）同時記錄此時的子列長度countnums= i - j + 1。
②：else if(A[i] < L)
  此時新加的可能情況為res = res + countnums;注意此時的A[i]實際上是比給定的範圍要小的點，是不滿足題意的，但是如果它之前的countnums不為0，說明最近一次的子列加上該點也可以構成新子列，但是必須以countnums中的子列（該子列中的最後一個元素為A[j]嚴格滿足A[j] >= L && A[j] <= R）中的某個元素為開始，以A[i]為結尾）。
③A[i]>R
  此時意味著一箇舊子列的結束，更新各個對應的引數

class Solution {
public:
    int numSubarrayBoundedMax(vector<int>& A, int L, int R) {
        int size = A.size();
        int countnums = 0;
        int res = 0;
        int j = 0;
        for(int i = 0; i<size; i++)
        {
            if(A[i] >= L && A[i] <= R)
            {
                res += i - j + 1;
                countnums = i - j + 1;
            }
            else if(A[i] < L)
            {
                res = res + countnums;
            }
            else
            {
                j = i + 1;
                countnums = 0;
                if(j > size)
                {
                    break;
                }
            }
        }
        return res;
    }
};