問題描述
On a 2D plane, we place stones at some integer coordinate points. Each coordinate point may have at most one stone.
Now, a move consists of removing a stone that shares a column or row with another stone on the grid.
What is the largest possible number of moves we can make?
Example 1:
Input: stones = [[0,0],[0,1],[1,0],[1,2],[2,1],[2,2]]
Output: 5
Example 2:
Input: stones = [[0,0],[0,2],[1,1],[2,0],[2,2]]
Output: 3
Example 3:
Input: stones = [[0,0]]
Output: 0
Note:
- 1 <= stones.length <= 1000
- 0 <= stones[i][j] < 10000
簡單翻譯一下,有一個二維矩陣,矩陣中有一些石塊,在同一行或者同一列的石塊我們可以remove掉,但是要至少保留其中一個,問在這個矩陣中最多remove掉多少石塊。
思路:
這道題和LC200–最大連通區域題有些類似,我們可以把在同一行/列的石塊看成connected,每個connected區域保留一個石塊,這樣的話最後用total stone - number of connected area
即爲最後結果。那麼怎麼計算 number of connected area呢?我寫了兩種方法:DFS和UnionFind供參考。
代碼:
//DFS
//Time : O(N^2)
//Space: O(N)
public int removeStones(int[][] stones) {
Set<int[]> visited = new HashSet<>();
int numOfIslands = 0;
for (int[] s1:stones){
if (!visited.contains(s1)){
dfs(s1, visited, stones);
numOfIslands++;
}
}
return stones.length - numOfIslands;
}
private void dfs(int[] s1, Set<int[]> visited, int[][] stones){
visited.add(s1);
for (int[] s2: stones){
if (!visited.contains(s2)){
if (s1[0] == s2[0] || s1[1] == s2[1])
dfs(s2, visited, stones);
}
}
}
UnionFind:
//UnionFind
Map<Integer, Integer> f = new HashMap<>();
int islands = 0;
public int removeStones1(int[][] stones) {
for (int i = 0; i < stones.length; ++i)
union(stones[i][0], ~stones[i][1]);
return stones.length - islands;
}
public int find(int x) {
if (f.putIfAbsent(x, x) == null)
islands++;
if (x != f.get(x))
f.put(x, find(f.get(x)));
return f.get(x);
}
public void union(int x, int y) {
x = find(x);
y = find(y);
if (x != y) {
f.put(x, y);
islands--;
}
}