描述
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
Note:
1 <= nums.length <= 10^3
nums.length == n
1 <= nums[i] <= 100
1 <= left <= right <= n * (n + 1) / 2
解析
根据题意，最暴利的解法就是将 nums 的子数组和都算出来放入 res 中，然后对 res 进行排序，最后将 res[left-1:right] 的子数组求和并对 10**9+7 取模即可，只是这种方法太没技术含量了，鄙视自己的投机取巧。
解答
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
"""
res = []
for i in range(n):
for j in range(i+1, n+1):
res.append(sum(nums[i:j]))
res.sort()
return sum(res[left-1:right])%(10**9+7)
运行结果
Runtime: 5696 ms, faster than 10.53% of Python online submissions for Range Sum of Sorted Subarray Sums.
Memory Usage: 33.7 MB, less than 36.84% of Pyt
原题链接：leetcode.com/problems/ra…
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