网站制作模板下载培训体系包括四大体系

Leetcode695. 岛屿的最大面积

文章链接：代码随想录
题目链接：695. 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

文章链接：代码随想录
题目链接：1020. 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

文章链接：代码随想录
题目链接：130. 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

文章链接：代码随想录
题目链接：417. 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

文章链接：代码随想录
题目链接：827.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！！！