LeetCode 2104. 子数组范围和（单调栈）

示例 1：
输入：nums = [1,2,3]
输出：4
解释：nums 的 6 个子数组如下所示：
[1]，范围 = 最大 - 最小 = 1 - 1 = 0 
[2]，范围 = 2 - 2 = 0
[3]，范围 = 3 - 3 = 0
[1,2]，范围 = 2 - 1 = 1
[2,3]，范围 = 3 - 2 = 1
[1,2,3]，范围 = 3 - 1 = 2
所有范围的和是 0 + 0 + 0 + 1 + 1 + 2 = 4

示例 2：
输入：nums = [1,3,3]
输出：4
解释：nums 的 6 个子数组如下所示：
[1]，范围 = 最大 - 最小 = 1 - 1 = 0
[3]，范围 = 3 - 3 = 0
[3]，范围 = 3 - 3 = 0
[1,3]，范围 = 3 - 1 = 2
[3,3]，范围 = 3 - 3 = 0
[1,3,3]，范围 = 3 - 1 = 2
所有范围的和是 0 + 0 + 0 + 2 + 0 + 2 = 4

示例 3：
输入：nums = [4,-2,-3,4,1]
输出：59
解释：nums 中所有子数组范围的和是 59
 
提示：
1 <= nums.length <= 1000
-10^9 <= nums[i] <= 10^9

class Solution:
    def subArrayRanges(self, nums: List[int]) -> int:
        def getRange(nums, neg):
            nums_ = copy.deepcopy(nums)
            if neg:
                for i in range(len(nums)):
                    nums_[i] = -nums_[i] # 找最小值的左右区间，取个负数，逻辑跟正数一样
            L, R = list(range(len(nums))), list(range(len(nums)))
            q = []
            for i in range(len(nums)):
                while len(q) and nums_[q[-1]] < nums_[i]:
                    q.pop()
                L[i] = q[-1]+1 if len(q) else 0
                q.append(i)
            q = []
            for i in reversed(range(len(nums))):
                while len(q) and nums_[q[-1]] <= nums_[i]:
                    q.pop() # 注意上下只能有一个 = 号
                R[i] = q[-1]-1 if len(q) else (len(nums)-1)
                q.append(i)
            return L, R
        L1, R1 = getRange(nums, neg=False)
        L2, R2 = getRange(nums, neg=True)
        ans = 0
        for i in range(len(nums)):
            # print(L1[i], R1[i], L2[i], R2[i])
            ans += (R1[i]-i+1)*(i-L1[i]+1)*nums[i]-(R2[i]-i+1)*(i-L2[i]+1)*nums[i]
        return ans