力扣每日一题2022-05-26困难题：掉落的方块

• • 题目描述
  • 思路
  • • 枚举
    • • Python实现
      • Java实现
      • C++

掉落的方块

用数组heights记录各个方块掉落后的高度。对于第i个掉落的方块，如果底部区间与第j个掉落的方块有重叠，则它掉落后的高度至少为heights[j]+sizei，其中j＜i且sizei为第i个掉落的方块的边长。因此对于第i个掉落的方块，heights[i]的初始值为sizei，暴力枚举所有之前掉落的方块，如果两个的底部区间有重叠，则更新heights[i]=max(heights[i], heights[j]+sizei)。
因为要返回一个所有已落稳的方块的最大堆叠高度列表，从i=1开始，不断更新heights，然后返回即可。

class Solution:
    def fallingSquares(self, positions: List[List[int]]) -> List[int]:
        n = len(positions)
        heights = [0] * n
        for i, (left1, side1) in enumerate(positions):
            right1 = left1 + side1 - 1
            heights[i] = side1
            for j in range(i):
                left2, right2 = positions[j][0], positions[j][0] + positions[j][1] - 1
                if right1 >= left2 and right2 >= left1:
                    heights[i] = max(heights[i], heights[j] + side1)
        for i in range(1, n):
            heights[i] = max(heights[i], heights[i-1])
        return heights

class Solution {
    public List fallingSquares(int[][] positions) {
        int n = positions.length;
        List heights = new ArrayList();
        for (int i = 0; i < n; i++) {
            int left1 = positions[i][0], right1 = positions[i][0]+positions[i][1]-1;
            int height = positions[i][1];
            for (int j = 0; j < i; j++) {
                int left2 = positions[j][0], right2 = positions[j][0] + positions[j][1] - 1;
                if (right1 >= left2 && right2 >= left1) {
                    height = Math.max(height, heights.get(j) + positions[i][1]);
                }
            }
            heights.add(height);
        }
        for (int i = 1; i < n; i++) {
            heights.set(i, Math.max(heights.get(i), heights.get(i-1)));
        }
        return heights;
    }
}

class Solution {
public:
    vector fallingSquares(vector>& positions) {
        int n = positions.size();
        vector heights(n);
        for (int i = 0; i < n; i++) {
            int left1 = positions[i][0], right1 = positions[i][0] + positions[i][1] - 1;
            heights[i] = positions[i][1];
            for (int j = 0; j < i; j++) {
                int left2 = positions[j][0], right2 = positions[j][0] + positions[j][1] - 1;
                if (right1 >= left2 && right2 >= left1) {
                    heights[i] = max(heights[i], heights[j] + positions[i][1]);
                }
            }
        }
        for (int i = 1; i < n; i++) {
            heights[i] = max(heights[i], heights[i - 1]);
        }
        return heights;
    }
};