LeetCode刷题笔记 130. 被围绕的区域

题目描述

给定一个二维的矩阵，包含 ‘X’ 和 ‘O’（字母 O）。

找到所有被 ‘X’ 围绕的区域，并将这些区域里所有的 ‘O’ 用 ‘X’ 填充。

示例:
X X X X
X O O X
X X O X
X O X X
运行你的函数后，矩阵变为：
X X X X
X X X X
X X X X
X O X X

总结

SDC1：DFS，想不到效率还挺高的，还是那句话，先实现，再优化。
SDC2：并查集，虽然效率不高，但真的是一种很好的思路。

Sample Code 1

class Solution {
    public void solve(char[][] board) {
        if (board.length == 0) return;
        int rows = board.length;
        int cols = board[0].length;
        
        for (int i = 0; i < cols; i++) {
            // if (board[0][i] == 'O') 
                dfs(0, i, board);            
            // if (board[board.length - 1][i] == 'O') 
                dfs(board.length - 1, i, board);
        }
        
        for (int i = 1; i < rows - 1; i++) {
            // if (board[i][0] == 'O')
                dfs(i, 0, board);
            // if (board[i][board[0].length - 1] == 'O')
                dfs(i, board[0].length - 1, board);
        }

        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (board[i][j] == '*') board[i][j] = 'O';
                else if(board[i][j] == 'O') board[i][j] = 'X';
            }
        }
    }

    private void dfs(int i, int j, char[][] board) {
        if (i < 0 || j < 0 || i == board.length || j == board[0].length) 
            return;
        
        if (board[i][j] == '*') return;
            
        if (board[i][j] == 'O') {
            board[i][j] = '*';
            dfs(i + 1, j, board);
            dfs(i, j + 1, board);
            dfs(i, j - 1, board);
            dfs(i - 1, j, board);
        }
    }
}

Sample Code 2

class Solution {
    int rows, cols;
    
    public void solve(char[][] board) {
        if(board == null || board.length == 0) return;
        rows = board.length;
        cols = board[0].length;
        
        UnionFind uf = new UnionFind(rows * cols + 1);
        int dummyNode = rows * cols;
        
        for(int i = 0; i < rows; i++) {
            for(int j = 0; j < cols; j++) {
                if(board[i][j] == 'O') {
                    if(i == 0 || i == rows-1 || j == 0 || j == cols-1) {
                        uf.union(dummyNode,node(i,j));
                    }else {
                        if(board[i-1][j] == 'O')  uf.union(node(i,j), node(i-1, j));
                        if(board[i+1][j] == 'O')  uf.union(node(i,j), node(i+1, j));
                        if(board[i][j-1] == 'O')  uf.union(node(i,j), node(i, j-1));
                        if(board[i][j+1] == 'O')  uf.union(node(i,j), node(i, j+1));
                    }
                }
            }
        }
        
        for(int i = 0; i < rows; i++) {
            for(int j = 0; j < cols; j++) {
                if(uf.isConnected(node(i,j), dummyNode)) {
                    board[i][j] = 'O';
                }else {
                    board[i][j] = 'X';
                }
            }
        }
    }
    
    int node(int i, int j) {
        return i * cols + j;
    }
}

class UnionFind {
    int [] parents;
    public UnionFind(int totalNodes) {
        parents = new int[totalNodes];
        for(int i = 0; i < totalNodes; i++) {
            parents[i] = i;
        }
    }
    
    void union(int node1, int node2) {
        int root1 = find(node1);
        int root2 = find(node2);
        if(root1 != root2) {
            parents[root2] = root1;
        }
    }
    
    int find(int node) {
        while(parents[node] != node) {
            parents[node] = parents[parents[node]];
            node = parents[node];
        }
        return node;
    }
    
    boolean isConnected(int node1, int node2) {
        return find(node1) == find(node2);
    }
}

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/surrounded-regions