这道题也比较基础，算是贪心吧？每一步把当前最大最后的归位就好。
自己的解
class Solution {
public:
vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) {
int now = people.size();
while (now != 0){
int biggest = 0;
for (int i = 0; i < now; i++){
if (people[i].first > people[biggest].first){
biggest = i;
}else if (people[i].first == people[biggest].first && people[i].second < people[biggest].second){
biggest = i;
}
}
--now;
swap(people[biggest], people[now]);
pair <int, int> temp = people[now];
for(int i = now; i <now + temp.second; i ++){
people[i] = people[i+1];
}
people[now + temp.second] = temp;
}
return people;
}
};
别人家的解。。。
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;
}
总结，多熟悉库函数T T
