leetcode 907. Sum of Subarray Minimums

Given an array of integers A, find the sum of min(B), where B ranges over every (contiguous) subarray of A.

Since the answer may be large, return the answer modulo 10^9 + 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 <= A.length <= 30000
1 <= A[i] <= 30000

分析：假设数组为a，对于index为i的数，假设其连续左边直到m的数都比他大或者相等，连续右边n个数都比他大，那以a[i]为最小值的子列的个数一共有m*n个，那其实对于数组中的每个数，我们只要计算出其对应的m和n就ok， m和n可以用一个单调栈就可以很容易实现。

class Solution(object):
    def sumSubarrayMins(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        if(len(A)==1):
            return A[0]
        left = [0 for i in range(len(A))]
        right = [0 for i in range(len(A))]
        left[0] = 1
        stack = []
        stack.append(0)
        for i in range(1, len(A)):
            while len(stack)>0 and A[i]<=A[stack[len(stack)-1]]:
                stack.pop()
            if len(stack)==0:
                left[i] = i+1
            else:
                left[i] = i-stack[len(stack)-1]
            stack.append(i)
        stack=[]    
        stack.append(len(A)-1)
        right[len(A)-1] = 1
        for i in range(len(A)-2, -1, -1):
            while len(stack)>0 and A[i]