LeetCode 2104. Sum of Subarray Ranges

原题链接在这里：https://leetcode.com/problems/sum-of-subarray-ranges/

题目：

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
• -109 <= nums[i] <= 109

题解：

Find sum(max - min) in all the subarrays. It is equal to find sum(max) - sum(min) in all the subarrays.

Using the way in  to find min and max of all the subarrays.

Time Complexity: O(n). n = nums.length.

Space: O(1).

AC Java:

 1 class Solution {
 2     public long subArrayRanges(int[] nums) {
 3         if(nums == null || nums.length == 0){
 4             return 0;
 5         }
 6         
 7         Stack stk = new Stack<>();
 8         long sum = 0;
 9         int n = nums.length;
10         for(int i = 0; i <= n; i++){
11             while(!stk.isEmpty() && nums[stk.peek()] > (i == n ? Integer.MIN_VALUE : nums[i])){
12                 int minInd = stk.pop();
13                 int pre = stk.isEmpty() ? -1 : stk.peek();
14                 sum -= (long)nums[minInd] * (i - minInd) * (minInd - pre);
15             }
16             
17             stk.push(i);
18         }
19         
20         stk.clear();
21         for(int i = 0; i <= n; i++){
22             while(!stk.isEmpty() && nums[stk.peek()] < (i == n ? Integer.MAX_VALUE : nums[i])){
23                 int maxInd = stk.pop();
24                 int pre = stk.isEmpty() ? -1 : stk.peek();
25                 sum += (long)nums[maxInd] * (i - maxInd) * (maxInd - pre);
26             }
27             
28             stk.push(i);
29         }
30         
31         return sum;
32     }
33 }

类似.