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]]

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：

解法一：

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;
    }
};

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，这种方法没有没有使用vector自带的insert或者erase函数，而是通过一个变量cnt和k的关系来将元素向前移动到正确位置，移动到方法是通过每次跟前面的元素交换位置，使用题目中给的例子来演示过程：

[[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]

排序后：

[[7,0], [7,1], [6,1], [5,0], [5,2], [4,4]]

交换顺序：

[[7,0], [6,1], [7,1], [5,0], [5,2], [4,4]]

[[5,0], [7,0], [6,1], [7,1], [5,2], [4,4]]

[[5,0], [7,0], [5,2], [6,1], [7,1], [4,4]]

[[5,0], [7,0], [5,2], [6,1], [4,4], [7,1]]

解法二：

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 = 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;
    }
};

下面这种解法跟解法一很相似，只不过没有使用额外空间，而是直接把位置不对的元素从原数组中删除，直接加入到正确的位置上，参见代码如下：

解法三：

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++) {
            auto p = people[i];
            if (p.second != i) {
                people.erase(people.begin() + i);
                people.insert(people.begin() + p.second, p);
            }
        }
        return people;
    }
};

类似题目：

Count of Smaller Numbers After Self

参考资料：

https://discuss.leetcode.com/topic/60394/easy-concept-with-python-c-java-solution/3

https://discuss.leetcode.com/topic/60413/short-java-solution-without-using-extra-space