【CODE】Queue Reconstruction by Height

406. Queue Reconstruction by Height

Medium

2111251Add to ListShare

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]]

class Solution {
public:
    static bool cmp(vector<int> a,vector<int> b){
        if(a[0]<b[0]) return false;
        if(a[0]>b[0]) return true;
        if(a[0]==b[0]){
            if(a[1]<b[1]) return true;
            else return false;
        }
        return true;
    }
    vector<vector<int>> reconstructQueue(vector<vector<int>>& people) {
        if(people.size()==0) return people;
        sort(people.begin(), people.end() , cmp);
        int max = people[0][0];
        for (int i = 0; i < people.size();i++){
            int a = people[i][0], b = people[i][1];
            if(people[i][0]<=max){
                int j = people[i][1];
                int k = 0;
                while(j--){
                    if(people[k][0]>=people[i][0]) k++;
                }
                for (int m = i-1; m >=k;m--){
                    people[m + 1][0] = people[m][0];
                    people[m + 1][1] = people[m][1];
                }
                people[k][0] = a;
                people[k][1] = b;
            }
            max = a; 
        }
        return people;
    }
};