LeetCode[Medium]------947. Most Stones Removed with Same Row or Column

问题描述

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. 1 <= stones.length <= 1000
2. 0 <= stones[i][j] < 10000

简单翻译一下，有一个二维矩阵，矩阵中有一些石块，在同一行或者同一列的石块我们可以remove掉，但是要至少保留其中一个，问在这个矩阵中最多remove掉多少石块。

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思路：

这道题和LC200–最大连通区域题有些类似，我们可以把在同一行/列的石块看成connected，每个connected区域保留一个石块，这样的话最后用total stone - number of connected area即为最后结果。那么怎么计算 number of connected area呢？我写了两种方法：DFS和UnionFind供参考。

代码:

	//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--;
        }
    }