【leetcode】1508. Range Sum of Sorted Subarray Sums

题目如下：

Given the array nums consisting of n positive integers. You computed the sum of all non-empty continous subarrays from the array and then sort them in non-decreasing order, creating a new array of n * (n + 1) / 2 numbers.

Return the sum of the numbers from index left to index right (indexed from 1), inclusive, in the new array. Since the answer can be a huge number return it modulo 10^9 + 7.

Example 1:

Input: nums = [1,2,3,4], n = 4, left = 1, right = 5
Output: 13 
Explanation: All subarray sums are 1, 3, 6, 10, 2, 5, 9, 3, 7, 4. After sorting them in non-decreasing order we have 
the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 1 to ri = 5 is 1 + 2 + 3 + 3 + 4 = 13. 

Example 2:

Input: nums = [1,2,3,4], n = 4, left = 3, right = 4
Output: 6
Explanation: The given array is the same as example 1. We have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of 
the numbers from index le = 3 to ri = 4 is 3 + 3 = 6.

Example 3:

Input: nums = [1,2,3,4], n = 4, left = 1, right = 10
Output: 50

Constraints:

• 1 <= nums.length <= 10^3
• nums.length == n
• 1 <= nums[i] <= 100
• 1 <= left <= right <= n * (n + 1) / 2

解题思路：怎么说呢，nums.length最大才1000，表示O(n^2)的时间复杂度完全可以接受，那就计算出所有的子数组的和吧。

代码如下：

class Solution(object):
    def rangeSum(self, nums, n, left, right):
        """
        :type nums: List[int]
        :type n: int
        :type left: int
        :type right: int
        :rtype: int
        """
        val = []
        for i in range(len(nums)):
            count = 0
            for j in range(i,len(nums)):
                count += nums[j]
                val.append(count)
        val.sort()
        return sum(val[left-1:right])