LeetCode406. Queue Reconstruction by Height

题目

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

这道题目是讲一队人排队，一组(h,k)代表一个人，h表示这个人的身高，k表示这个人前面有几个比他高或跟他一样高的人，按照这个规则重新排队。

我的思路是首先将这个数组中的每对pair按照第一个元素的逆序排序，如果第一个元素一样，就按第二个元素的顺序来排，得到一个新的数组。观察可以知道，遍历这个新数组时，将每个pair的第二个元素作为下标插入到新的数组的相应位置，这个位置就是前面比他高或一样高的人数，得到的新数组就是题目所要求的数组。

代码

class Solution {
public:
void insert_sort(vector<pair<int,int>>& people)
{
for(int i = 1;i < people.size();i++)
{
pair<int,int> temp = people[i];
int j = i;
//第一个元素逆序，或第一个元素相等时第二个元素顺序
while(j > 0 && (people[j-1].first < temp.first || (people[j-1].first == temp.first)&&(people[j-1].second > temp.second)))
{
people[j] = people[j-1];
j--;
}
people[j] = temp;
}
}
vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) {
insert_sort(people);
vector<pair<int, int>> res;
//第二个元素作为位置插入新数组中
for(int i = 0;i < people.size();i++)
{
res.insert(res.begin()+people[i].second,people[i]);
}
return res;
}
};