【LeetCode】406. Queue Reconstruction by Height

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406. Queue Reconstruction by Height

Description：
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.
Difficulty：Medium

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]]

方法：排序插空

• Time complexity : $O\left ( nlgn \right )$?
• Space complexity : $O\left ( 1 \right )$
  思路：
  按照身高从大到小排序，身高相同时，按照数量从小打大排序。
  按照排序好的数组顺序插入到空数组中即可，原因是先插入大的元素，后插入的小元素不会影响大元素的相对位置关系，并且插入时已经保证了自身的相对位置正确

input:  [[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]
sort :  [[7,0], [7,1], [6,1],  [5,0], [5,2], [4,4]]
step 1: [[7,0], [7,1]]
step 2: [[7,0], [6,1], [7,1]]
step 3: [[5,0], [7,0], [5,2], [6,1], [7,1]]

class Solution {
public:
    vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) {
        auto cmop = [](const pair<int, int>& p1, pair<int, int>& p2){
            return p1.first > p2.first || (p1.first == p2.first && p1.second < p2.second);
        };
        sort(people.begin(), people.end(), cmop);
        vector<pair<int, int>> res;
        for(auto& p : people)
            res.insert(res.begin() + p.second, p);
        return res;
    }
};