2104. 子数组范围和

给你一个整数数组 nums 。nums 中，子数组的 范围 是子数组中最大元素和最小元素的差值。

返回 nums 中 所有 子数组范围的 和 。

子数组是数组中一个连续 非空 的元素序列。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/sum-of-subarray-rang
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

import java.util.ArrayDeque;
import java.util.Deque;

class Solution {
    public long subArrayRanges(int[] nums) {
        int n = nums.length;
        int[] minLeft = new int[n];
        int[] minRight = new int[n];
        int[] maxLeft = new int[n];
        int[] maxRight = new int[n];
        Deque<Integer> minStack = new ArrayDeque<>();
        Deque<Integer> maxStack = new ArrayDeque<>();
        // 3 3 3 3
        for (int i = 0; i < n; i++) {
            while (!minStack.isEmpty() && nums[minStack.peek()] >= nums[i]) {
                minStack.pop();
            }
            minLeft[i] = minStack.isEmpty() ? -1 : minStack.peek();
            minStack.push(i);

            while (!maxStack.isEmpty() && nums[maxStack.peek()] <= nums[i]) {
                maxStack.pop();
            }
            maxLeft[i] = maxStack.isEmpty() ? -1 : maxStack.peek();
            maxStack.push(i);
        }
        minStack.clear();
        maxStack.clear();
        for (int i = n - 1; i >= 0; i--) {
            while (!minStack.isEmpty() && nums[minStack.peek()] > nums[i]) {
                minStack.pop();
            }
            minRight[i] = minStack.isEmpty() ? n : minStack.peek();
            minStack.push(i);

            while (!maxStack.isEmpty() && nums[maxStack.peek()] < nums[i]) {
                maxStack.pop();
            }
            maxRight[i] = maxStack.isEmpty() ? n : maxStack.peek();
            maxStack.push(i);
        }

        long sumMax = 0, sumMin = 0;
        for (int i = 0; i < n; i++) {
            sumMax += (long) (maxRight[i] - i) * (i - maxLeft[i]) * nums[i];
            sumMin += (long) (minRight[i] - i) * (i - minLeft[i]) * nums[i];
        }
        return sumMax - sumMin;
    }
}