[LeetCode] 1762. Buildings With an Ocean View

There are n buildings in a line. You are given an integer array heights of size n that represents the heights of the buildings in the line.

The ocean is to the right of the buildings. A building has an ocean view if the building can see the ocean without obstructions. Formally, a building has an ocean view if all the buildings to its right have a smaller height.

Return a list of indices (0-indexed) of buildings that have an ocean view, sorted in increasing order.

Example 1:

Input: heights = [4,2,3,1]
Output: [0,2,3]
Explanation: Building 1 (0-indexed) does not have an ocean view because building 2 is taller.

Example 2:

Input: heights = [4,3,2,1]
Output: [0,1,2,3]
Explanation: All the buildings have an ocean view.

Example 3:

Input: heights = [1,3,2,4]
Output: [3]
Explanation: Only building 3 has an ocean view.

Example 4:

Input: heights = [2,2,2,2]
Output: [3]
Explanation: Buildings cannot see the ocean if there are buildings of the same height to its right.

Constraints:

• 1 <= heights.length <= 105
• 1 <= heights[i] <= 109

我這里提供兩種思路，一是貪心，二是單調棧。

因為海是在數組的右邊，數組最后一個元素是沒有被擋住的，所以它有oceanview。照着這個思路，我們從數組的最右邊開始往左掃描，如果遇到更高的樓，那么這個更高的樓就是有oceanview的；反之則是沒有。

時間O(n)

空間O(n)

Java實現

class Solution {
    public int[] findBuildings(int[] heights) {
        int max = 0;
        List<Integer> list = new ArrayList<>();
        for (int i = heights.length - 1; i >= 0; i--) {
            if (heights[i] > max) {
                list.add(i);
                max = heights[i];
            }
        }

        int[] res = new int[list.size()];
        Collections.reverse(list);
        for (int i = 0; i < res.length; i++) {
            res[i] = list.get(i);
        }
        return res;
    }
}

 

單調棧其實也是類似的思路，但是單調棧可以讓我們從左往右遍歷input數組。我們這里做的是一個單調遞增的棧。

時間O(n)

空間O(n)

Java實現

class Solution {
    public int[] findBuildings(int[] heights) {
        Deque<Integer> stack = new ArrayDeque<>();
        for (int i = 0; i < heights.length; i++) {
            while (!stack.isEmpty() && heights[stack.peekLast()] <= heights[i]) {
                stack.pollLast();
            }
            stack.addLast(i);
        }
        int n = stack.size();
        int[] res = new int[n];
        while (!stack.isEmpty()) {
            res[--n] = stack.pollLast();
        }
        return res;
    }
}

 

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