图论第二天|695. 岛屿的最大面积 1020. 飞地的数量 130. 被围绕的区域 417. 太平洋大西洋水流问题 827.最大人工岛

Leetcode695. 岛屿的最大面积

思路：dfs

class Solution {
public:
    int count;
    int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};
    void dfs(vector<vector<int>>& grid, vector<vector<bool>>& visited, int x, int y){
        for (int i = 0; i < 4; i++){
            int nex = x + dir[i][0];
            int ney = y + dir[i][1];

            if (nex < 0 || nex >= grid.size() || ney < 0 || ney >= grid[0].size()) continue;
            if (!visited[nex][ney] && grid[nex][ney] == 1){
                visited[nex][ney] = true;
                count++;
                dfs(grid, visited, nex, ney);
            }
        }
    }


    int maxAreaOfIsland(vector<vector<int>>& grid) {
        int result = 0;
        vector<vector<bool>> visited(grid.size(), vector<bool>(grid[0].size(), 0));
        for (int i = 0; i < grid.size(); i++){
            for (int j = 0; j < grid[0].size(); j++){
                if (!visited[i][j] && grid[i][j] == 1){
                    visited[i][j] = true;
                    count = 1;
                    dfs (grid, visited, i, j);
                    result = max(result, count);
                }
            }
        }
        return result;
    }

};

bfs

class Solution {
public:
    int count;
    int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};
    void bfs(vector<vector<int>>& grid, vector<vector<bool>>& visited, int x, int y){
        queue<pair<int, int>> que;
        que.push({x, y});
        while(!que.empty()){
            pair<int, int> cur = que.front();
            que.pop();
            for (int i = 0; i < 4; i++){
                int nex = cur.first + dir[i][0];
                int ney = cur.second + dir[i][1];

                if (nex < 0 || nex >= grid.size() || ney < 0 || ney >= grid[0].size()) continue;
                if (!visited[nex][ney] && grid[nex][ney] == 1){
                    visited[nex][ney] = true;
                    count++;
                    que.push({nex, ney});
                }
            }
        }
        
    }


    int maxAreaOfIsland(vector<vector<int>>& grid) {
        int result = 0;
        vector<vector<bool>> visited(grid.size(), vector<bool>(grid[0].size(), 0));
        for (int i = 0; i < grid.size(); i++){
            for (int j = 0; j < grid[0].size(); j++){
                if (!visited[i][j] && grid[i][j] == 1){
                    visited[i][j] = true;
                    count = 1;
                    bfs (grid, visited, i, j);
                    result = max(result, count);
                }
            }
        }
        return result;
    }

};

Leetcode1020. 飞地的数量

思路：dfs

class Solution {
public:
    int count = 0;
    int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};
    void dfs(vector<vector<int>>& grid, int x, int y){
        grid[x][y] = 0;
        count++;
        for (int i = 0; i < 4; i++){
            int nex = x + dir[i][0];
            int ney = y + dir[i][1];

            if (nex < 0 || nex >= grid.size() || ney < 0 || ney >= grid[0].size()) continue;
            if (grid[nex][ney] == 1) dfs(grid, nex, ney);
        }
        return ;
    }
    
    
    int numEnclaves(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        int result = 0;

        for (int i = 0; i < m; i++){
            if(grid[i][0] == 1) dfs(grid, i, 0);
            if(grid[i][n - 1] == 1) dfs(grid, i, n - 1);
        }

        for (int j = 0; j < n; j++){
            if (grid[0][j] == 1) dfs(grid, 0, j);
            if (grid[m - 1][j] == 1) dfs(grid, m - 1, j);
        }

        for (int i = 0; i < m; i++){
            for (int j = 0; j < n; j++){
                if (grid[i][j] == 1) {
                    count = 0;
                    dfs(grid, i, j);
                    result += count;
                }
            }
        }
        return result;
    }
};

Leetcode130. 被围绕的区域

思路：dfs

class Solution {
public:
    int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};
    void dfs(vector<vector<char>>& board, int x, int y){
        board[x][y] = 'A';
        for (int i = 0; i < 4; i++){
            int nex = x + dir[i][0];
            int ney = y + dir[i][1];

            if (nex < 0 || nex >= board.size() || ney < 0 || ney >= board[0].size()) continue;
            if (board[nex][ney] == 'O') dfs(board, nex, ney);
        }
    }    
    
    void solve(vector<vector<char>>& board) {
        int m = board.size();
        int n = board[0].size();

        for (int i = 0; i < m; i++){
            if (board[i][0] == 'O') dfs(board, i, 0);
            if (board[i][n - 1] == 'O') dfs(board, i, n - 1);
        }

        for (int j = 0; j < n; j++){
            if (board[0][j] == 'O') dfs(board, 0, j);
            if (board[m - 1][j] == 'O') dfs(board, m - 1, j);
        }

        for (int i = 0; i < m; i++){
            for (int j = 0; j < n; j++){
                if (board[i][j] == 'O') board[i][j] = 'X';
                if (board[i][j] == 'A') board[i][j] = 'O';
            }
        }
    }
};

Leetcode417. 太平洋大西洋水流问题

思路：注意终止条件 if (visited[x][y]) return ;

class Solution {
public:
    int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};
    void dfs(vector<vector<int>>& heights, vector<vector<bool>>& visited, int x, int y){
        // 注意终止条件
        if (visited[x][y]) return ;
        visited[x][y] = true;
        for (int i = 0; i < 4; i++){
            int nex = x + dir[i][0];
            int ney = y + dir[i][1];

            if (nex < 0 || nex >= heights.size() || ney < 0 || ney >= heights[0].size()) continue;
            if (heights[x][y] <= heights[nex][ney]) dfs(heights, visited, nex, ney);
        }
    }
    
    
    
    vector<vector<int>> pacificAtlantic(vector<vector<int>>& heights) {
        int m = heights.size();
        int n = heights[0].size();
        vector<vector<bool>> pacific(m, vector<bool>(n, false));
        vector<vector<bool>> atlantic(m, vector<bool>(n, false));
        vector<vector<int>> result;

        for (int i = 0; i < m; i++){
            dfs(heights, pacific, i, 0);
            dfs(heights, atlantic, i, n - 1);
        }

        for (int j = 0; j < n; j++){
            dfs(heights, pacific, 0, j);
            dfs(heights, atlantic, m - 1, j);
        }

        for (int i = 0; i < m; i++){
            for (int j = 0; j < n; j++){
                if (pacific[i][j] && atlantic[i][j]) result.push_back({i, j});
            }
        }
        
        return result;
    }
};

Leetcode827.最大人工岛

思路：dfs，先用map记录原有的每块陆地的大小，再在0处遍历连接陆地，选择最大值。

class Solution {
public:
    int count;
    int mark = 2;
    int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};
    
    void dfs(vector<vector<int>>& grid, vector<vector<bool>>& visited, int x, int y){
        if (visited[x][y] || grid[x][y] == 0) return ;
        visited[x][y] = true;
        count++;
        grid[x][y] = mark;
        for (int i = 0; i < 4; i++){
            int nex = x + dir[i][0];
            int ney = y + dir[i][1];

            if (nex < 0 || nex >= grid.size() || ney < 0 || ney >= grid[0].size()) continue;
            dfs(grid, visited, nex, ney);
        }
    }
    
    
    
    int largestIsland(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        vector<vector<bool>> visited(m, vector<bool>(n, false));
        unordered_map<int, int> gridNum;
        bool isAllGrid = true;
        int result = 0;

        for (int i = 0; i < m; i++){
            for (int j = 0; j < n; j++){
                if (grid[i][j] == 0) isAllGrid = false;
                if (!visited[i][j] && grid[i][j] == 1){
                    count = 0;
                    dfs(grid, visited, i, j);
                    gridNum[mark] = count;
                    mark++;
                }
            }
        }
        // cout << count << endl;
        // cout << gridNum[2] << endl;

        if (isAllGrid) return n * m;

        unordered_set<int> visitedGrid;

        for (int i = 0; i < m; i++){
            for (int j = 0; j < n; j++){
                visitedGrid.clear();
                if (grid[i][j] == 0){
                    count = 1;
                    for (int k = 0; k < 4; k++){
                        int nex = i + dir[k][0];
                        int ney = j + dir[k][1];

                        if (nex < 0 || nex >= grid.size() || ney < 0 || ney >= grid[0].size()) continue;
                        if (visitedGrid.count(grid[nex][ney]) == 0){
                            count += gridNum[grid[nex][ney]];
                            visitedGrid.insert(grid[nex][ney]);
                        }
                    }
                    result = max(result, count);
                }
            }
        }

        return result;
    }
};

图论第二天打卡，整体来说套路感挺重的，理解和做起来挺简单的，但写多了也头晕哈哈，加油！！！