406.QueueReconstructionbyHeight.h

/*
2016-10-13
bluepp
May the force be with me!

https://leetcode.com/problems/queue-reconstruction-by-height/

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]]
*/

    vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) {
        
        sort(people.begin(), people.end(), [](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>> res;
        for (auto p : people)
        {
            res.insert(res.begin()+p.second, p);
        }
        
        return res;
    }


/* no extra space */
    vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) {
        
        sort(people.begin(), people.end(), [](const pair<int, int>& a, const pair<int, int>& b) {
            return a.first > b.first || (a.first == b.first && a.second < b.second);
        });
        
        for (int i = 1; i < people.size(); ++i) {
            int cnt = 0;
            for (int j = 0; j < i; ++j) {
                if (cnt == people[i].second) {
                    pair<int, int> t = people[i];
                    for (int k = i - 1; k >= j; --k) {
                        people[k + 1] = people[k];
                    }
                    people[j] = t;
                    break;
                }
                if (people[j].first >= people[i].first) ++cnt;
            }
        }
        
        return people;

    }