2104. 子数组范围和 :「区间 DP」&「递推」&「单调栈」

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
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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
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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
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class Solution {
    public long subArrayRanges(int[] nums) {
        int n = nums.length;
        int[][][] f = new int[n][n][2];
        for (int i = 0; i < n; i++) f[i][i][0] = f[i][i][1] = nums[i];
        for (int len = 2; len <= n; len++) {
            for (int l = 0; l + len - 1 < n; l++) {
                int r = l + len - 1;
                f[l][r][0] = Math.min(nums[r], f[l][r - 1][0]);
                f[l][r][1] = Math.max(nums[r], f[l][r - 1][1]);
            }
        }
        long ans = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                ans += f[i][j][1] - f[i][j][0];
            }
        }
        return ans;
    }
}
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class Solution {
    public long subArrayRanges(int[] nums) {
        int n = nums.length;
        long ans = 0;
        for (int i = 0; i < n; i++) {
            int min = nums[i], max = nums[i];
            for (int j = i + 1; j < n; j++) {
                min = Math.min(min, nums[j]);
                max = Math.max(max, nums[j]);
                ans += max - min;
            }
        }
        return ans;
    }
}
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class Solution {
    int n;
    public long subArrayRanges(int[] nums) {
        n = nums.length;
        long[] min = getCnt(nums, true), max = getCnt(nums, false);
        long ans = 0;
        for (int i = 0; i < n; i++) ans += (max[i] - min[i]) * nums[i];
        return ans;
    }
    long[] getCnt(int[] nums, boolean isMin) {
        int[] a = new int[n], b = new int[n];
        Deque<Integer> d = new ArrayDeque<>();
        for (int i = 0; i < n; i++) {
            while (!d.isEmpty() && (isMin ? nums[d.peekLast()] >= nums[i] : nums[d.peekLast()] <= nums[i])) d.pollLast();
            a[i] = d.isEmpty() ? -1 : d.peekLast();
            d.addLast(i);
        }
        d.clear();
        for (int i = n - 1; i >= 0; i--) {
            while (!d.isEmpty() && (isMin ? nums[d.peekLast()] > nums[i] : nums[d.peekLast()] < nums[i])) d.pollLast();
            b[i] = d.isEmpty() ? n : d.peekLast();
            d.addLast(i);
        }
        long[] ans = new long[n];
        for (int i = 0; i < n; i++) ans[i] = (i - a[i]) * 1L * (b[i] - i);
        return ans;
    }
}
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