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.
Pick out tallest group of people and sort them in a subarray (S). Since there's no other groups of people taller than them, therefore each guy's index will be just as same as his k value.
For 2nd tallest group (and the rest), insert each one of them into (S) by k value. So on and so forth.
E.g.
input: [[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]
subarray after step 1: [[7,0], [7,1]]
subarray after step 2: [[7,0], [6,1], [7,1]]
...
It's not the most concise code, but I think it well explained the concept.
class Solution(object):
def reconstructQueue(self, people):
if not people: return []
# obtain everyone's info
# key=height, value=k-value, index in original array
peopledct, height, res = {}, [], []
for i in xrange(len(people)):
p = people[i]
if p[0] in peopledct:
peopledct[p[0]] += (p[1], i),
else:
peopledct[p[0]] = [(p[1], i)]
height += p[0],
height.sort() # here are different heights we have
# sort from the tallest group
for h in height[::-1]:
peopledct[h].sort()
for p in peopledct[h]:
res.insert(p[0], people[p[1]])
return res
但速度不快
class Solution {
public:
vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) {
list<tuple<int,int,int>> tuples;
for(int i=0;i<people.size();i++){
tuples.push_back(make_tuple(people[i].first,people[i].second,people[i].second));
}
vector<pair<int,int>> res;
while(tuples.size()>0){
//find minimum with 0
int minh=-1;
list<tuple<int,int,int>>::iterator it_del;
for(auto it=tuples.begin();it!=tuples.end();++it){
if(get<2>(*it)==0){
if(get<0>(*it)<minh || minh==-1){
minh=get<0>(*it);
it_del=it;
}
}
}
res.push_back(make_pair(minh,get<1>(*it_del)));
tuples.erase(it_del);
//update rest
for(auto& it:tuples){
if(get<0>(it)<=minh){
get<2>(it)=get<2>(it)-1;
}
}
}
return res;
}
};