406. Queue Reconstruction by Height (Medium)

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]]

Solution 1: Greedy

这道题给了我们一个队列，队列中的每个元素是一个pair，分别为身高和前面身高不低于当前身高的人的个数，让我们重新排列队列，使得每个pair的第二个参数都满足题意。首先我们来看一种超级简洁的方法，不得不膜拜想出这种解法的大神。首先我们给队列先排个序，按照身高高的排前面，如果身高相同，则第二个数小的排前面。然后我们新建一个空的数组，遍历之前排好序的数组，然后根据每个元素的第二个数字，将其插入到res数组中对应的位置，参见代码如下：

class Solution {
public:
    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);});
        vector<pair<int, int>> res;
        for (auto a: people) {
            res.insert(res.begin()+a.second,a);
        }
        return res;
    }
};

Solution 2: O(1) extra space

上面那种方法是简洁，但是用到了额外空间，我们来看一种不使用额外空间的解法，

class Solution {
public:
    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 = 0; 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;
    }
};