Question

1762. Buildings With an Ocean View

中文文档

Description

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.

 

Constraints:

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

Solutions

Solution 1

class Solution:
  def findBuildings(self, heights: List[int]) -> List[int]:
    ans = []
    mx = 0
    for i in range(len(heights) - 1, -1, -1):
      if heights[i] > mx:
        ans.append(i)
        mx = heights[i]
    return ans[::-1]

class Solution {
  public int[] findBuildings(int[] heights) {
    int n = heights.length;
    List<Integer> ans = new ArrayList<>();
    int mx = 0;
    for (int i = heights.length - 1; i >= 0; --i) {
      if (heights[i] > mx) {
        ans.add(i);
        mx = heights[i];
      }
    }
    Collections.reverse(ans);
    return ans.stream().mapToInt(Integer::intValue).toArray();
  }
}

class Solution {
public:
  vector<int> findBuildings(vector<int>& heights) {
    vector<int> ans;
    int mx = 0;
    for (int i = heights.size() - 1; ~i; --i) {
      if (heights[i] > mx) {
        ans.push_back(i);
        mx = heights[i];
      }
    }
    reverse(ans.begin(), ans.end());
    return ans;
  }
};

func findBuildings(heights []int) (ans []int) {
  mx := 0
  for i := len(heights) - 1; i >= 0; i-- {
    if v := heights[i]; v > mx {
      ans = append(ans, i)
      mx = v
    }
  }
  for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 {
    ans[i], ans[j] = ans[j], ans[i]
  }
  return
}

function findBuildings(heights: number[]): number[] {
  const ans: number[] = [];
  let mx = 0;
  for (let i = heights.length - 1; ~i; --i) {
    if (heights[i] > mx) {
      ans.push(i);
      mx = heights[i];
    }
  }
  return ans.reverse();
}

/**
 * @param {number[]} heights
 * @return {number[]}
 */
var findBuildings = function (heights) {
  const ans = [];
  let mx = 0;
  for (let i = heights.length - 1; ~i; --i) {
    if (heights[i] > mx) {
      ans.push(i);
      mx = heights[i];
    }
  }
  return ans.reverse();
};