LeetCodeアルゴリズム問題集-406.Queue Reconstruction by Height(高再構築キュー経由)
ランダムなキューがあると仮定すると、各人は1対の数字(h,k)として記述され、hはその人の高さであり、kはその人の前にあり、h以上の人の数である.以上の規則に従ってこのキューを再構築する必要がある.
英語原文:
Suppose you have a random list of people standing in a queue. Each person is described by a pair of integers
注:1100未満の人数
例:
回答:
最も効率的なバージョン:
英語原文:
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:1100未満の人数
例:
:
[[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]
:
[[5,0], [7,0], [5,2], [6,1], [4,4], [7,1]]
回答:
最も効率的なバージョン:
struct node{
pairval;
int count;
node*left, *right;
node(pairn)
{
val = n;
count = 0;
left = NULL, right = NULL;
}
};
class Solution {
public:
vector> reconstructQueue(vector>& people) {
auto cmp = [](paira, pairb){return a.first==b.first?a.secondb.first;};
sort(people.begin(), people.end(), cmp);
vector>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, pairval)
{
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> tra(node* root)
{
vector>ans;
stacksta;
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;
}
};
最も簡潔なバージョン:class Solution {
public:
vector> reconstructQueue(vector>& people) {
auto comp = [](const pair& p1, const pair& p2)
{ return p1.first > p2.first || (p1.first == p2.first && p1.second < p2.second); };
sort(people.begin(), people.end(), comp);
vector> res;
for (auto& p : people)
res.insert(res.begin() + p.second, p);
return res;
}
};