406. Queue Reconstruction by Height

最新推荐文章于 2023-11-18 01:25:37 发布

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最新推荐文章于 2023-11-18 01:25:37 发布

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分类专栏： leetcode 需回顾贪心算法

本文链接：https://blog.youkuaiyun.com/qq_34038762/article/details/78291053

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本文介绍了一种解决LeetCode上队列重建问题的方法。该问题要求根据每个人的身高及在其前方比他高的人数来重建队列。文章提供了一个算法，首先对输入数据进行排序，然后将每个元素插入到合适的位置。

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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]]

我的解答：
先按照身高h从大到小排序，相同身高的按k从小到大排序。
[7,0] [7,1] [6,1] [5,0] [5,2] [4,4]
取出开头的元素，插入下面的队列中k个元素的后面。
[5,0] [7,0] [5,2] [6,1] [4,4] [7,1]

bool compare(pair<int, int> &a, pair<int, int> &b) {
    if (a.first > b.first)
        return true;
    else if (a.first == b.first && a.second< b.second)
        return true;
    else return false;
}
class Solution {
public:
    vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) {
        sort(people.begin(), people.end(), compare);
        vector<pair<int, int>> result;
        for(auto i: people) {
            result.insert(result.begin() + i.second, i);
        }
        return result;
    }
};