題目
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:
1Input: heights = [4,2,3,1]
2Output: [0,2,3]
3Explanation: Building 1 (0-indexed) does not have an ocean view because building 2 is taller.
Example 2:
1Input: heights = [4,3,2,1]
2Output: [0,1,2,3]
3Explanation: All the buildings have an ocean view.
Example 3:
1Input: heights = [1,3,2,4]
2Output: [3]
3Explanation: Only building 3 has an ocean view.
Constraints:
$$
\begin{array}{l}
1 <= heights.length <= 10^5 \\
1 <= heights[i] <= 10^9
\end{array}
$$
我的思路
我們現在要看有多少建築物可以看到右邊的海洋，所以當我在第 i 個建築物時，i 建築物右邊的全部建築物都需要比 i 建築物矮。
所以我會從最右開始往左看，用一個變數去記錄當下最高建築物的高度，如果當下的建築物沒有比記錄的建築物高，也就代表可以看到海洋。
由於回傳的結果必須是由左到右的，所以這邊我會先用 Stack 將可看見海洋的建築物存下來，最後在將 Stack 裡面的東西轉換回 Array 返回。
程式碼
1class Solution {
2 public int[] findBuildings(int[] heights) {
3 // 初始化最右邊的建築物
4 int heightest = heights[heights.length - 1];
5 Stack<Integer> stack = new Stack();
6 // 最右邊的放子一定看的到海
7 stack.push(heights.length - 1);
8
9 for(int i = heights.length - 1; i >= 0; i--) {
10 if (heights[i] > heightest) {
11 stack.push(i);
12 heightest = heights[i];
13 }
14 }
15
16 int size = stack.size();
17 int[] result = new int[size];
18 for(int i = 0; i < size; i++) {
19 result[i] = stack.pop();
20 }
21 return result;
22 }
23}
Time complexity: O(N)
Space complexity: O(N)
評論