Leetcode 406. Queue Reconstruction by Height

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题目描述

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.

输入：
输出：排序后的队伍
如：

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]]

思路1.

先对输入的序列对进行排序，首先根据高度排序，高度大的在前面；其次，当高度相同时，k小的在前。上述排序过后结果为：[[7,0],[7,1],[6,1],[5,0],[5,2],[4,4]]。随后，组成序列。基本思路是，大的数字先插入，插入位置为k。这样，能保证在插入第（i,k）时，已有序列本身都>=i，这是插到第k个位置是合理的。这种贪心策略每一步都是合理的。

代码

class Solution {public:    static bool comp(pair<int, int>& p1, pair<int, int>& p2){                            return p1.first > p2.first || (p1.first == p2.first && p1.second < p2.second);     }    vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) {        vector<pair<int, int>> res;        sort(people.begin(), people.end(), this->comp);        for (int i =0;i<people.size();i++) {            pair<int, int>& p = people[i];            res.insert(res.begin() + p.second, p);}        return res;    }};