网站建设后的专人维护,wordpress的安装步骤,如何建网站费用多少,工业设计和产品设计的区别目录 Leetcode695. 岛屿的最大面积Leetcode1020. 飞地的数量Leetcode130. 被围绕的区域Leetcode417. 太平洋大西洋水流问题Leetcode827.最大人工岛 Leetcode695. 岛屿的最大面积 文章链接#xff1a;代码随想录 题目链接#xff1a;695. 岛屿的最大面积 思路#xff1a;dfs … 目录 Leetcode695. 岛屿的最大面积Leetcode1020. 飞地的数量Leetcode130. 被围绕的区域Leetcode417. 太平洋大西洋水流问题Leetcode827.最大人工岛 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(vectorvectorint grid, vectorvectorbool 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(vectorvectorint grid) {int result 0;vectorvectorbool visited(grid.size(), vectorbool(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(vectorvectorint grid, vectorvectorbool visited, int x, int y){queuepairint, int que;que.push({x, y});while(!que.empty()){pairint, 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(vectorvectorint grid) {int result 0;vectorvectorbool visited(grid.size(), vectorbool(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(vectorvectorint 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(vectorvectorint 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(vectorvectorchar 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(vectorvectorchar 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(vectorvectorint heights, vectorvectorbool 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);}}vectorvectorint pacificAtlantic(vectorvectorint heights) {int m heights.size();int n heights[0].size();vectorvectorbool pacific(m, vectorbool(n, false));vectorvectorbool atlantic(m, vectorbool(n, false));vectorvectorint 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(vectorvectorint grid, vectorvectorbool 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(vectorvectorint grid) {int m grid.size();int n grid[0].size();vectorvectorbool visited(m, vectorbool(n, false));unordered_mapint, 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_setint 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;}
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