leetcode 406. 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]]

题目大意：给定输入序列对，重新排列，使得对于第i个对[h,k]他前面有k个大于等于h的对。

思路：先按照h从大到小w从小到大排序，这样能保证最少的修改次数，即只用把当前的对，放到对应的位置就行了。

class Solution {
public:
    vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) {
        sort(people.begin(), people.end(), [](pair<int,int> a, pair<int,int>b){
            if (a.first == b.first) return a.second < b.second;
            return a.first > b.first;
        });
        vector<pair<int,int> > ans;
        for (int i = 0; i < people.size(); ++i) {
            if (people[i].second < i) {
                ans.insert(ans.begin()+people[i].second, people[i]);
            }
            else ans.push_back(people[i]);
        }
        return ans;
    }
};