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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:
<pre>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. </pre>
Example 2:
<pre>Input: heights = [4,3,2,1] Output: [0,1,2,3] Explanation: All the buildings have an ocean view. </pre>
Example 3:
<pre>Input: heights = [1,3,2,4] Output: [3] Explanation: Only building 3 has an ocean view. </pre>
Constraints:
1 <= heights.length <= 10<sup>5</sup>1 <= heights[i] <= 10<sup>9</sup>
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题目大意：
大海在右边，求看到大海的大厦的下标
解题思路：
数组元素之间大小关系且保持顺序，用stack
解题步骤：
N/A
注意事项：
Python代码：
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7def findBuildings(self, heights: List[int]) -> List[int]:
stack = []
for i in range(len(heights)):
while stack and heights[i] >= heights[stack[-1]]:
stack.pop()
stack.append(i) # 4 3 1
return stack
算法分析：
时间复杂度为O(n)，空间复杂度O(n)
算法II解题思路(推荐)：
储水或者低谷题用向左向右的局部最值法。这题只需要单边最值
解题步骤：
类似于Leetcode 42的trapping rain water，看不到大海表示在低位。此题只求单边，从右往左扫描一次。
注意事项：
Python代码：
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7def findBuildings2(self, heights: List[int]) -> List[int]:
right_max, res = 0, []
for i in reversed(range(len(heights))):
if heights[i] > right_max:
res.append(i)
right_max = max(right_max, heights[i])
return res[::-1]
算法分析：
时间复杂度为O(n)，空间复杂度O(1)
