Leetcode 406. Queue Reconstruction by Height

-题目-
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]]

-思路-
这是一道贪心算法的题目，每次加入一个人，使得当前队伍的重排满足条件，这样最终加入的人也能使整个队伍满足重排的条件。为了使贪心算法有最大效益，需要对输入的原队伍进行排序，由题意可知，矮的人对高的人的排序不产生影响，因此先从高的人开始重排。然后将身高矮的人不断加进前面的人已产生好的排列。需要注意的是，当身高相同时，需要再根据前面的更高的人数按照从小到大的顺序进行排序，因为前面更高人数比较小的人会对前面更高的人数比较大的人产生影响。（因为它可能插到前面更高人数比较大的人前面，从而导致错误）

-代码-

class Solution {public:    static bool cmp1(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>> reconstructQueue(vector<pair<int, int>>& people) {        vector<pair<int, int> > newQueue;         //按身高和前面比他高的人排序        sort(people.begin(), people.end(), cmp1);         for(int i = 0; i < people.size() ; i++) {            cout << people[i].first << " " << people[i].second << endl;         }        if(people.size() == 0) return newQueue;         newQueue.push_back(people[0]);         vector<pair<int, int>>::iterator i, j;         for(i = people.begin()+1; i != people.end(); i++) {            int s = i -> second;             bool isin = 0;             for(j = newQueue.begin(); j != newQueue.end(); j++) {                if(!s) {                    newQueue.insert(j, *i);                     isin = 1;                     break;                 }                else if(j->first >= i->first) s--;             }            if(!isin) newQueue.insert(newQueue.end(), *i);         }        return newQueue;     }};