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; } };
本文转自博客园Grandyang的博客，原文链接：根据高度重建队列[LeetCode] Queue Reconstruction by Height ，如需转载请自行联系原博主。
