200. 岛屿数量(BFS+并查集)

200. 岛屿数量

给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。

岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。

此外，你可以假设该网格的四条边均被水包围。

 

示例 1：

输入：grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
输出：1

示例 2：

输入：grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
输出：3

 

提示：

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 300
• grid[i][j] 的值为 '0' 或 '1'

解法一：（BFS）

class Solution {
public:
    static constexpr int g_direciton[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
    void bfs(const vector<vector<char>> &grid, int x, int y, vector<vector<bool>> &visited) {
        int row = grid.size();
        int col = grid[0].size();
        queue<std::pair<int, int>> q;
        q.push(make_pair(x, y));
        visited[x][y] = true;
        while (!q.empty()) {
            int size = q.size();
            for (int i = 0; i < size; i++) {
                std::pair<int, int> tmp = q.front();
                q.pop();
                for (int j = 0; j < 4; j++) {
                    int xNext = tmp.first + g_direciton[j][0];
                    int yNext = tmp.second + g_direciton[j][1];
                    if ((xNext >= 0 && xNext < row) && (yNext >= 0 && yNext < col) && 
                        !visited[xNext][yNext] && (grid[xNext][yNext] == '1')) {
                        q.push(make_pair(xNext, yNext));
                        visited[xNext][yNext] = true;
                    }
                }
            }
        }
    }
    int numIslands(vector<vector<char>>& grid) {
        if (grid.size() == 0 || grid[0].size() == 0) {
            return 0;
        }
        int row = grid.size();
        int col = grid[0].size();
        vector<vector<bool>> visited(row, vector<bool>(col, false));
        int cnt = 0;
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                if (grid[i][j] == '0') {
                    continue;
                }
                if (visited[i][j]) {
                    continue;
                }
                bfs(grid, i, j, visited);
                cnt++;
            }
        }
        return cnt;
    }
};

 

解法二：（并查集）

class UnionFind {
public:
    void initUnionFind(vector<vector<char>> &grid) {
        int row = grid.size();
        int col = grid[0].size();
        int n = row * col;
        parent.resize(n);
        cnt = 0;
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                int index = i * col + j; // 二维转一维
                parent[index] = index;
                if (grid[i][j] == '1') {
                    cnt++; // 初始时岛屿的数量为陆地数量(数值为1)
                }
            }
        }
    }
    int findRoot(int x) {
        if (parent[x] == x) {
            return parent[x];
        }
        parent[x] = findRoot(parent[x]); // 递归找到x的根节点
        return parent[x];
    }
    void unify(int x, int y) {
        int xRoot = findRoot(x);
        int yRoot = findRoot(y);
        if (xRoot == yRoot) {
            return;
        }
        parent[xRoot] = yRoot;
        cnt--;
    }
    int getConnectedCnt() {
        return cnt;
    }
private:
    vector<int> parent;
    int cnt; // 连通分量数
};
class Solution : UnionFind {
public:
    int numIslands(vector<vector<char>>& grid) {
        if (grid.size() == 0 || grid[0].size() == 0) {
            return 0;
        }
        initUnionFind(grid);
        int row = grid.size();
        int col = grid[0].size();
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                int index = i * col + j;
                if (grid[i][j] == '1') {
                    // 左右为1的岛屿合并
                    if ((i + 1 < row) && grid[i + 1][j] == '1') {
                        unify(index, (i + 1) * col + j);
                    }
                    // 上下为1的岛屿合并
                    if ((j + 1 < col) && grid[i][j + 1] == '1') {
                        unify(index, i * col + j + 1);
                    }
                }
            }
        }
        return getConnectedCnt();
    }
};