[LeetCode 406] Queue Reconstruction by Height

最新推荐文章于 2020-10-30 15:18:28 发布

ExcitedZhang

最新推荐文章于 2020-10-30 15:18:28 发布

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分类专栏： LeetCode 文章标签： LeetCode 数组

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本文详细解析了一种基于高度和前排高者数量的队列重构算法，通过具体实例展示了算法的实现过程与效率，适用于人群排队场景的模拟。

Suppose you have a random list of people standing in a queue. Each person is described by a pair of integers (h, k), where h is the height of the person and k is the number of people in front of this person who have a height greater than or equal to h. Write an algorithm to reconstruct the queue.

Note:
The number of people is less than 1,100.

Example

Input:
[[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]

Output:
[[5,0], [7,0], [5,2], [6,1], [4,4], [7,1]]

分析

这一题只需要理清一个思路就可以：一定按照身高从高到低放置。因此高的人的站位决定着低的人站位。

以题目的例子为例，我们首先放置高度为7的，按照k从小到大的次序放置

[7,0] [7,1]

然后放置身高为6的，这个时候6的位置一定是确定的

[7,0] [6,1] [7,1]

再放置身高为5的，先放[5,0]，再放[5,2]，放置的位置就是第k个位置

[5,0] [7,0] [5,2] [6,1] [7,1]

一次类推，即可得到所有的排列。

Code

class Solution {
public:
    static bool cmp(pair<int, int>& a, pair<int, int>& b)
    {
        return a.first < b.first || (a.first == b.first && a.second > b.second);
    }
    
    vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) {
        vector<pair<int, int>> res;
        sort(people.begin(), people.end(), cmp);
        
        for (int i = people.size()-1; i >= 0; i--)
        {
            res.insert(res.begin() + people[i].second,
                      people[i]);
        }
        return res;
    }
};

运行效率

Runtime: 40 ms, faster than 95.92% of C++ online submissions for Queue Reconstruction by Height.

Memory Usage: 10.3 MB, less than 85.56% of C++ online submissions for Queue Reconstruction by Height.