[Leetcode] 699. Falling Squares 解题报告

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原创最新推荐文章于 2026-09-20 16:36:07 发布 687 阅读 AI 写同款· GEO 优化 ›

Leetcode 699. Falling Squares Leetcode 699. Falling Squares 很久没有写题解了，主要是leetcode没啥有意思的题目，写来没啥意思。今天这题还算有点意思，可以分析分析。描述给你一个序列表示下落的俄罗斯方块的情况。序列里面含有 NNN 对值 (L,S)(L,S)(L, S)，其中LLL表示方块下落的左坐标，SSS表示下落方块的边长，这里下落的方块永远是正方形。其中N≤1000,L≤... 阅读详情

题目：

On an infinite number line (x-axis), we drop given squares in the order they are given.

The i-th square dropped (positions[i] = (left, side_length)) is a square with the left-most point being positions[i][0] and sidelength positions[i][1].

The square is dropped with the bottom edge parallel to the number line, and from a higher height than all currently landed squares. We wait for each square to stick before dropping the next.

The squares are infinitely sticky on their bottom edge, and will remain fixed to any positive length surface they touch (either the number line or another square). Squares dropped adjacent to each other will not stick together prematurely.

Return a list ans of heights. Each height ans[i] represents the current highest height of any square we have dropped, after dropping squares represented by positions[0], positions[1], ..., positions[i].

Example 1:

Input: [[1, 2], [2, 3], [6, 1]]
Output: [2, 5, 5]
Explanation:

After the first drop of positions[0] = [1, 2]:
_aa
_aa
-------
The maximum height of any square is 2.

After the second drop of positions[1] = [2, 3]:
__aaa
__aaa
__aaa
_aa__
_aa__
--------------
The maximum height of any square is 5.  
The larger square stays on top of the smaller square despite where its center
of gravity is, because squares are infinitely sticky on their bottom edge.

After the third drop of positions[1] = [6, 1]:
__aaa
__aaa
__aaa
_aa
_aa___a
--------------
The maximum height of any square is still 5.

Thus, we return an answer of [2, 5, 5].

Example 2:

Input: [[100, 100], [200, 100]]
Output: [100, 100]
Explanation: Adjacent squares don't get stuck prematurely - only their bottom edge can stick to surfaces.

Note:

1 <= positions.length <= 1000.
1 <= positions[i][0] <= 10^8.

1 <= positions[i][1] <= 10^6.

思路：

有点类似于“城市天际线”那道题目，没怎么看懂网上参考题目的复杂逻辑，后面再补充具体解释吧。。。

代码：

class Solution {
public:
    vector<int> fallingSquares(vector<pair<int, int>>& positions) {
        map<pair<int,int>, int> mp;
        mp[{0,1000000000}] = 0;                 // the initialization
        vector<int> ans;
        int mx = 0;
        for (auto &v : positions) {
            vector<vector<int>> toAdd;
            int len = v.second;
            int a = v.first, b = v.first + v.second, h = 0;
            auto it = mp.upper_bound({a,a});    // the first ont that is larger than a
            if (it != mp.begin() && (--it)->first.second <= a) {
                ++it;
            }
            while (it != mp.end() && it->first.first < b) {
                if (it->first.first < a) {
                    toAdd.push_back({it->first.first, a, it->second});
                }
                if (it->first.second > b) {
                    toAdd.push_back({b, it->first.second, it->second});
                }
                h = max(h, it->second);
                it = mp.erase(it);
            }
            mp[{a, b}] = h + len;
            for (auto &t : toAdd) {
                mp[{t[0], t[1]}] = t[2];
            }
            mx = max(mx, h + len);
            ans.push_back(mx);
        }
        return ans;
    }
};

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