LeetCode算法题集-406. Queue Reconstruction by Height（通过高重建队列）

假设你有条随机队列，每个人被描述为一对数字（h,k），其中h是该人的高，k是在该人前面且高大于等于h的人的数量。要求按以上规则去重建这条队列。

英语原文：

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.

注:

人数少于 1,100.

例子：

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

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

解答：

效率最高的版本：

struct node{
pair<int, int>val;
int count;
node*left, *right;
node(pair<int, int>n)
{
val = n;
count = 0;
left = NULL, right = NULL;
}
};
class Solution {
public:
vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) {
auto cmp = [](pair<int, int>a, pair<int, int>b){return a.first==b.first?a.second<b.second:a.first>b.first;};
sort(people.begin(), people.end(), cmp);
vector<pair<int,int>>ans;
node* root = NULL;
for(auto temp:people)
{
insert(root, temp.second+1, temp);
}
ans = tra(root);
return ans;
}
void insert(node* &root, int n, pair<int, int>val)
{
if(!root)root = new node(val);
else if(n>(root->count+1))insert(root->right, n - root->count-1, val);
else root->count++, insert(root->left, n, val);
}
vector<pair<int, int>> tra(node* root)
{
vector<pair<int, int>>ans;
stack<node*>sta;
while(!sta.empty()||root)
{
if(root)
{
sta.push(root);
root = root->left;
}
else
{
root = sta.top();
sta.pop();
ans.push_back(root->val);
root = root->right;
}
}
return ans;
}
};

最简洁的版本：

class Solution {
public:
vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) {
auto comp = [](const pair<int, int>& p1, const pair<int, int>& p2)
{ return p1.first > p2.first || (p1.first == p2.first && p1.second < p2.second); };
sort(people.begin(), people.end(), comp);
vector<pair<int, int>> res;
for (auto& p : people)
res.insert(res.begin() + p.second, p);
return res;
}
};