标签：nums int Sum stk Ranges range subarray Subarray sum
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 Sum of Subarray Minimums 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<Integer> 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 }
标签：nums,int,Sum,stk,Ranges,range,subarray,Subarray,sum 来源： https://www.cnblogs.com/Dylan-Java-NYC/p/16392013.html
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