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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，
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; } };
LeetCode All in One 题目讲解汇总(持续更新中...)
[LeetCode] Queue Reconstruction by Height 根据高度重建队列
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原文地址：http://www.cnblogs.com/grandyang/p/5928417.html
