You are given an integer array
nums
. The range of a subarray of nums
is the difference between the largest and smallest element in the subarray.Return the sum of all subarray ranges of
nums
.A subarray is a contiguous non-empty sequence of elements within an array.
Example 1:
Input: nums = [1,2,3]
Output: 4
Explanation: The 6 subarrays of nums are the following:
[1], range = largest - smallest = 1 - 1 = 0
[2], range = 2 - 2 = 0
[3], range = 3 - 3 = 0
[1,2], range = 2 - 1 = 1
[2,3], range = 3 - 2 = 1
[1,2,3], range = 3 - 1 = 2
So the sum of all ranges is 0 + 0 + 0 + 1 + 1 + 2 = 4.
Example 2:
Input: nums = [1,3,3]
Output: 4
Explanation: The 6 subarrays of nums are the following:
[1], range = largest - smallest = 1 - 1 = 0
[3], range = 3 - 3 = 0
[3], range = 3 - 3 = 0
[1,3], range = 3 - 1 = 2
[3,3], range = 3 - 3 = 0
[1,3,3], range = 3 - 1 = 2
So the sum of all ranges is 0 + 0 + 0 + 2 + 0 + 2 = 4.
Example 3:
Input: nums = [4,-2,-3,4,1]
Output: 59
Explanation: The sum of all subarray ranges of nums is 59.
Constraints:
1 <= nums.length <= 1000
-10<sup>9</sup> <= nums[i] <= 10<sup>9</sup>
题目大意:
求所有子数组的最大值最小值之差的和
Stack算法思路:
参考Leetcode 907,分别求子数组最小值的相反数,子数组的最大值,这两个值的和即为所求
注意事项:
- 最小值用递增栈,最大值用递减栈
Python代码:
1 | def subArrayRanges(self, nums: List[int]) -> int: |
算法分析:
时间复杂度为O(n)
,空间复杂度O(n)
累计和算法II解题思路:
比暴力法稍优,两重循环覆盖所有子数组[i, j],每轮循环得到最大最小值,然后O(1)内求该区间内所有最大最小值差值和。
Python代码:
1 | def subArrayRanges(self, nums: List[int]) -> int: |
算法分析:
时间复杂度为O(n^2)
,空间复杂度O(1)