In a group of N people (labelled 0, 1, 2, ..., N-1
), each person has different amounts of money, and different levels of quietness.
For convenience, we'll call the person with label x
, simply "person x
".
We'll say that richer[i] = [x, y]
if person x
definitely has more money than person y
. Note that richer
may only be a subset of valid observations.
Also, we'll say quiet[x] = q
if person x has quietness q
.
Now, return answer
, where answer[x] = y
if y
is the least quiet person (that is, the person y
with the smallest value of quiet[y]
), among all people who definitely have equal to or more money than person x
.
Example 1:
Input: richer = [[1,0],[2,1],[3,1],[3,7],[4,3],[5,3],[6,3]], quiet = [3,2,5,4,6,1,7,0]
Output: [5,5,2,5,4,5,6,7]
Explanation:
answer[0] = 5.
Person 5 has more money than 3, which has more money than 1, which has more money than 0.
The only person who is quieter (has lower quiet[x]) is person 7, but
it isn't clear if they have more money than person 0.
answer[7] = 7.
Among all people that definitely have equal to or more money than person 7
(which could be persons 3, 4, 5, 6, or 7), the person who is the quietest (has lower quiet[x])
is person 7.
The other answers can be filled out with similar reasoning.
Note:
1 <= quiet.length = N <= 500
0 <= quiet[i] < N
, allquiet[i]
are different.0 <= richer.length <= N * (N-1) / 2
0 <= richer[i][j] < N
richer[i][0] != richer[i][1]
richer[i]
's are all different.- The observations in
richer
are all logically consistent.
Runtime: 120 ms, faster than 8.56% of C++ online submissions for Loud and Rich.
class Solution {
public:
vector<int> loudAndRich(vector<vector<int>>& richer, vector<int>& quiet) {
int n = quiet.size();
vector<int> ret(n,0);
for(int i=0; i<n; i++) ret[i] = i;
vector<unordered_set<int>> graph(quiet.size());
vector<unordered_set<int>> revgraph(quiet.size());
for(int i=0; i<richer.size(); i++){
graph[richer[i][0]].insert(richer[i][1]);
}
for(int i=0; i<richer.size(); i++){
revgraph[richer[i][1]].insert(richer[i][0]);
}
queue<int> q;
for(int i=0; i<n; i++){
if(revgraph[i].empty()) q.push(i);
}
while(!q.empty()){
int node = q.front(); q.pop();
for(auto it=graph[node].begin(); it!=graph[node].end(); it++){
int childnode = *it;
ret[childnode] = quiet[ret[node]] > quiet[ret[childnode]] ? ret[childnode] : ret[node];
revgraph[childnode].erase(node);
if(revgraph[childnode].empty()) q.push(childnode);
}
}
return ret;
}
};
还有一种dfs的做法,速度要快很多,因为没有用到字典删除。而且还有这一句话
auto __ = []() {std::ios::sync_with_stdio(false); return 0;}();
auto __ =[]() { std::ios::sync_with_stdio(false); cin.tie(nullptr); return nullptr; }();
void dfs(vector<vector<int>> & graph,vector<int> & go,vector<int> & retv,vector<int>quiets,int index)
{
int quiet = quiets[index];
go[index] = 1;
retv[index] = index;
for(int i=0;i<graph[index].size();i++)
{
if(go[graph[index][i]])
{
int person = retv[graph[index][i]];
if(quiet>quiets[person])
{
quiet = quiets[person];
retv[index] = person;
}
}
else
{
dfs(graph,go,retv,quiets,graph[index][i]);
int person = retv[graph[index][i]];
if(quiet>quiets[person])
{
quiet = quiets[person];
retv[index] = person;
}
}
}
}
class Solution {
public:
vector<int> loudAndRich(vector<vector<int>>& richer, vector<int>& quiet) {
vector<vector<int>> graph(quiet.size(),vector<int>());
if(quiet.size() == 0)
return vector<int>();
for(auto ele:richer)
{
graph[ele[1]].push_back(ele[0]);
}
vector<int> go(quiet.size(),0);
vector<int> retv(quiet.size(),0);
for(int i =0;i<quiet.size();i++)
{
if(go[i]==0)
{
dfs(graph,go,retv,quiet,i);
}
}
return retv;
}
};