Given an array of integersA, find the sum ofmin(B), whereBranges overevery (contiguous) subarray ofA.

Since the answer may be large,return the answer modulo10^9 + 7.

Example 1:

Input: [3,1,2,4]
Output: 17
Explanation: Subarrays are [3], [1], [2], [4], [3,1], [1,2], [2,4], [3,1,2], [1,2,4], [3,1,2,4]. 
Minimums are 3, 1, 2, 4, 1, 1, 2, 1, 1, 1. Sum is 17.
 

Note:

1. 1 <= A.length <= 30000
2. 1 <= A[i] <= 30000

https://leetcode.com/problems/sum-of-subarray-minimums/discuss/178876/stack-solution-with-very-detailed-explanation-step-by-step

class Solution {
    public int sumSubarrayMins(int[] A) {
        int len = A.length;
        Stack<Integer> stack = new
 
 Stack<>();
        int[] left = new int[len];
        int[] right = new int[len];
        for(int i = 0; i < A.length; i++) {
            left[i] = i + 1;
            right[i] = len - i;
        }
        // previous less element
        for(int i = 0; i < len; i++){
            while(!stack.isEmpty() && A[stack.peek()] > A[i]) {
                stack.pop();
            }
            left[i] 
 
= stack.isEmpty() ? i + 1 : i - stack.peek();
            stack.push(i);
        }
        //next less element
        stack = new Stack<>();
        for(int i = 0; i < len; i++){
            while(!stack.isEmpty() && A[stack.peek()] > A[i]) {
                right[stack.peek()] = i - stack.peek();
                stack.pop();
            }
            stack.push(i);
        }
        int ans = 0;
        int mod = (int)1e9 + 7;
        for(int i = 0; i < len; i++) {
            ans = (ans + A[i] * left[i] * right[i]) % mod;
        }
        return ans;
    }
}

https://www.cnblogs.com/grandyang/p/8887985.html

monotonic stack