[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]]

 

 思路：

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：

 

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>> ret;
         for (auto a : people)
         {
             ret.insert(ret.begin() + a.second, a);
         }
         return ret;
     }

参考：

http://www.cnblogs.com/grandyang/p/5928417.html

转载于:https://www.cnblogs.com/hellowooorld/p/7090201.html