[leetcode] 1508. Range Sum of Sorted Subarray Sums

[leetcode] 1508. Range Sum of Sorted Subarray Sums

Description

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 = 5Output: 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 = 4Output: 6Explanation: 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 = 10Output: 50

Constraints:

1 <= nums.length <= 10^3nums.length == n1 <= nums[i] <= 1001 <= left <= right <= n * (n + 1) / 2

分析

题目的意思是：给你一个数组，求出其字数组的和，构成新的数组，然后排序取left到right区间的数的和。这道题我一开始想的是暴力破解，没想到暴力破解也能够ac，让我想不到，我看了一下其他人的做法，有的用到了数组和来计算，但是好像维护right大小的堆才是最优解，哈哈哈，有兴趣的人先研究一下哈。

代码

class Solution: def rangeSum(self, nums: List[int], n: int, left: int, right: int) -> int: arr=[] for i in range(n): t=0 for j in range(i,n): t+=nums[j] arr.append(t) arr.sort() res=0 for i in range(left-1,right): res+=arr[i] return res%(10**9 + 7)

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