LeetCode795. 區間子數組個數

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795. Number of Subarrays with Bounded Maximum

[Medium] 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].

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題目：給定一個元素都是正整數的數組A ，正整數 L 以及 R (L <= R)。求連續、非空且其中最大元素滿足大於等於L 小於等於R的子數組個數。

思路：雙指針，參考link。cnt表示以A[i]結尾的且滿足條件的子數組數組。因此分爲三種情況：A[i]落在[L,R]區間內，那麼滿足條件的子數組個數爲i-j+1；如果A[i] < L，那麼其子數組個數由上一個狀態的cnt決定；如果A[i] > R，那麼沒有以A[i]結尾的子數組能滿足條件。

工程代碼下載

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