代码随想录第六十二天——岛屿数量，岛屿的最大面积，飞地的数量
leetcode 200. 岛屿数量
题目链接：岛屿数量
本题思路，是用遇到一个没有遍历过的节点陆地，计数器就加一，然后把该节点陆地所能遍历到的陆地都标记上。在遇到标记过的陆地节点和海洋节点的时候直接跳过。 这样计数器就是最终岛屿的数量
方法一：深度优先搜索
版本一：
class Solution {
private:
int dir[4][2] = {0,1,1,0,-1,0,0,-1}; //四个方向
void dfs(vector<vector<char>>& grid, vector<vector<bool>>& visited, int x,int y){
for(int i=0;i<4;i++){
int nextx = x + dir[i][0];
int nexty = y + dir[i][1];
if(nextx<0 || nextx>=grid.size() || nexty<0 || nexty>=grid[0].size()) continue;
if(!visited[nextx][nexty] && grid[nextx][nexty]=='1'){//没有访问过的 同时 是陆地的
visited[nextx][nexty]=true;
dfs(grid,visited,nextx,nexty);
}
}
}
public:
int numIslands(vector<vector<char>>& grid) {
int n = grid.size(), m = grid[0].size();
vector<vector<bool>> visited = vector<vector<bool>>(n, vector<bool>(m, false));
int res=0;
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
if(!visited[i][j] && grid[i][j]=='1'){
visited[i][j] = true;
res++; // 遇到没访问过的陆地，+1
dfs(grid, visited, i, j); // 将与其链接的陆地都标记上 true
}
}
}
return res;
}
};
版本二：
// 版本二
class Solution {
private:
int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1}; //四个方向
void dfs(vector<vector<char>>& grid, vector<vector<bool>>& visited, int x, int y) {
if (visited[x][y] || grid[x][y] == '0') return; //终止条件：访问过的节点或者遇到海水
visited[x][y] = true; //标记访问过
for (int i = 0; i < 4; i++) {
int nextx = x + dir[i][0];
int nexty = y + dir[i][1];
if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue; //越界了，直接跳过
dfs(grid, visited, nextx, nexty);
}
}
public:
int numIslands(vector<vector<char>>& grid) {
int n = grid.size(), m = grid[0].size();
vector<vector<bool>> visited = vector<vector<bool>>(n, vector<bool>(m, false));
int res = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (!visited[i][j] && grid[i][j] == '1') {
res++; //遇到没访问过的陆地，+1
dfs(grid, visited, i, j); //将与其链接的陆地都标记上 true
}
}
}
return result;
}
};
方法二：广度优先搜索
只要加入队列就代表走过，就需要标记，而不是从队列拿出来的时候再去标记走过。
class Solution {
private:
int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1}; // 四个方向
void bfs(vector<vector<char>>& grid, vector<vector<bool>>& visited, int x, int y) {
queue<pair<int, int>> que;
que.push({x, y});
visited[x][y] = true; //只要加入队列，立刻标记
while(!que.empty()) {
pair<int ,int> cur = que.front(); que.pop();
int curx = cur.first;
int cury = cur.second;
for (int i = 0; i < 4; i++) {
int nextx = curx + dir[i][0];
int nexty = cury + dir[i][1];
if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue; // 越界了，直接跳过
if (!visited[nextx][nexty] && grid[nextx][nexty] == '1') {
que.push({nextx, nexty});
visited[nextx][nexty] = true; // 只要加入队列立刻标记
}
}
}
}
public:
int numIslands(vector<vector<char>>& grid) {
int n = grid.size(), m = grid[0].size();
vector<vector<bool>> visited = vector<vector<bool>>(n, vector<bool>(m, false));
int res= 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (!visited[i][j] && grid[i][j] == '1') {
res++; // 遇到没访问过的陆地，+1
bfs(grid, visited, i, j); //将与其链接的陆地都标记上true
}
}
}
return re;
}
};
leetcode 695. 岛屿的最大面积
方法一：深度优先搜索
版本一：
class Solution {
private:
int count;
int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1}; //四个方向
void dfs(vector<vector<int>>& grid, vector<vector<bool>>& visited, int x, int y) {
for (int i = 0; i < 4; i++) {
int nextx = x + dir[i][0];
int nexty = y + dir[i][1];
if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue; //越界了直接跳过
if (!visited[nextx][nexty] && grid[nextx][nexty] == 1) { //没有访问过的，同时是陆地的
visited[nextx][nexty] = true;
count++;
dfs(grid, visited, nextx, nexty);
}
}
}
public:
int maxAreaOfIsland(vector<vector<int>>& grid) {
int n = grid.size(), m = grid[0].size();
vector<vector<bool>> visited = vector<vector<bool>>(n, vector<bool>(m, false));
int res = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (!visited[i][j] && grid[i][j] == 1) {
count = 1; // 因为dfs处理下一个节点，所以这里遇到陆地了就先计数，dfs处理接下来的相邻陆地
visited[i][j] = true;
dfs(grid, visited, i, j); // 将与其链接的陆地都标记上 true
res = max(res, count);
}
}
}
return res;
}
};
版本二：
class Solution {
private:
int count;
int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1}; // 四个方向
void dfs(vector<vector<int>>& grid, vector<vector<bool>>& visited, int x, int y) {
if (visited[x][y] || grid[x][y] == 0) return; // 终止条件：访问过的节点 或者 遇到海水
visited[x][y] = true; // 标记访问过
count++;
for (int i = 0; i < 4; i++) {
int nextx = x + dir[i][0];
int nexty = y + dir[i][1];
if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue; // 越界了，直接跳过
dfs(grid, visited, nextx, nexty);
}
}
public:
int maxAreaOfIsland(vector<vector<int>>& grid) {
int n = grid.size(), m = grid[0].size();
vector<vector<bool>> visited = vector<vector<bool>>(n, vector<bool>(m, false));
int res = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (!visited[i][j] && grid[i][j] == 1) {
count = 0; // 因为dfs处理当前节点，所以遇到陆地计数为0，进dfs之后在开始从1计数
dfs(grid, visited, i, j); // 将与其链接的陆地都标记上 true
res = max(res, count);
}
}
}
return res;
}
};
方法二：广度优先搜索
class Solution {
private:
int count;
int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1}; //四个方向
void bfs(vector<vector<int>>& grid, vector<vector<bool>>& visited, int x, int y) {
queue<int> que;
que.push(x);
que.push(y);
visited[x][y] = true; //加入队列就意味节点是陆地可到达的点
count++;
while(!que.empty()) {
int xx = que.front();que.pop();
int yy = que.front();que.pop();
for (int i = 0 ;i < 4; i++) {
int nextx = xx + dir[i][0];
int nexty = yy + dir[i][1];
if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue; //越界
if (!visited[nextx][nexty] && grid[nextx][nexty] == 1) { // 节点没有被访问过且是陆地
visited[nextx][nexty] = true;
count++;
que.push(nextx);
que.push(nexty);
}
}
}
}
public:
int maxAreaOfIsland(vector<vector<int>>& grid) {
int n = grid.size(), m = grid[0].size();
vector<vector<bool>> visited = vector<vector<bool>>(n, vector<bool>(m, false));
int res = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (!visited[i][j] && grid[i][j] == 1) {
count = 0;
bfs(grid, visited, i, j); // 将与其链接的陆地都标记上 true
res = max(res, count);
}
}
}
return res;
}
};
leetcode 1020. 飞地的数量
题目链接：飞地的数量
本题要求找到不靠边的陆地面积，只要从周边找到陆地然后通过dfs或者bfs将周边靠陆地且相邻的陆地都变成海洋，然后再去重新遍历地图，统计此时还剩下的陆地。
方法一：深度优先搜索
class Solution {
private:
int dir[4][2] = {-1, 0, 0, -1, 1, 0, 0, 1}; //保存四个方向
int count; //统计符合题目要求的陆地空格数量
void dfs(vector<vector<int>>& grid, int x, int y) {
grid[x][y] = 0;
count++;
for (int i = 0; i < 4; i++) { // 向四个方向遍历
int nextx = x + dir[i][0];
int nexty = y + dir[i][1];
//超过边界
if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue;
//不符合条件，不继续遍历
if (grid[nextx][nexty] == 0) continue;
dfs (grid, nextx, nexty);
}
return;
}
public:
int numEnclaves(vector<vector<int>>& grid) {
int n = grid.size(), m = grid[0].size();
//从左侧边，和右侧边向中间遍历
for (int i = 0; i < n; i++) {
if (grid[i][0] == 1) dfs(grid, i, 0);
if (grid[i][m - 1] == 1) dfs(grid, i, m - 1);
}
//从上边和下边向中间遍历
for (int j = 0; j < m; j++) {
if (grid[0][j] == 1) dfs(grid, 0, j);
if (grid[n - 1][j] == 1) dfs(grid, n - 1, j);
}
count = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (grid[i][j] == 1) dfs(grid, i, j);
}
}
return count;
}
};
方法二：广度优先搜索
class Solution {
private:
int count = 0;
int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1}; // 四个方向
void bfs(vector<vector<int>>& grid, int x, int y) {
queue<pair<int, int>> que;
que.push({x, y});
grid[x][y] = 0; //只要加入队列，立刻标记
count++;
while(!que.empty()) {
pair<int ,int> cur = que.front(); que.pop();
int curx = cur.first;
int cury = cur.second;
for (int i = 0; i < 4; i++) {
int nextx = curx + dir[i][0];
int nexty = cury + dir[i][1];
if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue; // 越界了，直接跳过
if (grid[nextx][nexty] == 1) {
que.push({nextx, nexty});
count++;
grid[nextx][nexty] = 0; //只要加入队列立刻标记
}
}
}
}
public:
int numEnclaves(vector<vector<int>>& grid) {
int n = grid.size(), m = grid[0].size();
//从左侧边，和右侧边向中间遍历
for (int i = 0; i < n; i++) {
if (grid[i][0] == 1) bfs(grid, i, 0);
if (grid[i][m - 1] == 1) bfs(grid, i, m - 1);
}
//从上边和下边向中间遍历
for (int j = 0; j < m; j++) {
if (grid[0][j] == 1) bfs(grid, 0, j);
if (grid[n - 1][j] == 1) bfs(grid, n - 1, j);
}
count = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (grid[i][j] == 1) bfs(grid, i, j);
}
}
return count;
}
};