There are n rooms labeled from 0 to n - 1 and all the rooms are locked except for room 0. Your goal is to visit all the rooms. However, you cannot enter a locked room without having its key.
When you visit a room, you may find a set of distinct keys in it. Each key has a number on it, denoting which room it unlocks, and you can take all of them with you to unlock the other rooms.
Given an array rooms where rooms[i] is the set of keys that you can obtain if you visited room i, return true if you can visit all the rooms, or false otherwise.
Example 1:
Input: rooms = [[1],[2],[3],[]] Output: true Explanation: We visit room 0 and pick up key 1. We then visit room 1 and pick up key 2. We then visit room 2 and pick up key 3. We then visit room 3. Since we were able to visit every room, we return true.
Example 2:
Input: rooms = [[1,3],[3,0,1],[2],[0]] Output: false Explanation: We can not enter room number 2 since the only key that unlocks it is in that room.
Constraints:
n == rooms.length2 <= n <= 10000 <= rooms[i].length <= 10001 <= sum(rooms[i].length) <= 30000 <= rooms[i][j] < nrooms[i] are unique.class Solution {
public:
bool canVisitAllRooms(vector<vector<int>>& rooms) {
bool res = true;
vector<bool> vis(rooms.size(), false);
queue <int> q;
q.push(0);
vis[0] = true;
while(!q.empty()){
int cur = q.front();
for (int key: rooms[cur]){
if(vis[key])
continue;
else{
q.push(key);
vis[key] = true;
}
}
q.pop();
}
for_each(vis.begin(), vis.end(), [&res](bool b){
res &= b;
});
return res;
}
};class Solution {
public:
int threeSumClosest(vector<int> &num, int target) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
sort(num.begin(), num.end());
int n = num.size();
int ans = num[0] + num[1] + num[2];
for (int i = 0; i < n - 2; i++)
{
if (i > 0 && num[i] == num[i - 1]) continue;
int l = i + 1, r = n - 1;
int goal = target - num[i];
while (l < r)
{
while (l < r && num[l] + num[r] < goal)
{
if (abs(goal - num[l] - num[r]) < abs(ans - target))
ans = num[l] + num[r] + num[i];
l++;
}
while (l < r && num[l] + num[r] > goal)
{
if (abs(goal - num[l] - num[r]) < abs(ans - target))
ans = num[l] + num[r] + num[i];
r--;
}
if (l < r && num[l] + num[r] == goal)
{
ans = num[i] + num[l] + num[r];
l++, r--;
break;
}
}
if (ans == target) break;
}
return ans;
}
};