200. 岛屿数量 - 力扣（LeetCode）

class Solution {public int numIslands(char[][] grid) {int n = grid.length;//grid行数int m = grid[0].length;//grid列数int res = 0;for(int r = 0;r<n;r++){for(int c=0;c<m;c++){if(grid[r][c]=='1'){dfs(grid,r,c);res++;}}}return res;    }public void dfs(char[][] grid,int r,int c){if(!inArea(grid,r,c)){return;}if(grid[r][c]!='1'){return;}grid[r][c] = '2';dfs(grid,r+1,c);dfs(grid,r-1,c);dfs(grid,r,c+1);dfs(grid,r,c-1);}public boolean inArea(char[][] grid,int r,int c){return r>=0&&r<grid.length&&c>=0&&c<grid[0].length;}}

463. 岛屿的周长 - 力扣（LeetCode）

class Solution {public int islandPerimeter(int[][] grid) {for(int r = 0;r<grid.length;r++){for(int c = 0;c<grid[0].length;c++){if(grid[r][c] == 1){//题目说只有一个return dfs(grid,r,c);}}}return 0;}int dfs(int[][] grid,int r,int c){//函数因为坐标(r,c)超出网格范围,返回对应一条黄色的边if(!inArea(grid,r,c)){return 1;}//函数因为[当前格子是海洋格子]返回,对应一条蓝色的边if(grid[r][c]==0){return 1;}//函数因为[当前格子已经是遍历的陆地格子]返回,跟周长没有关系if(grid[r][c]!=1){return 0;}grid[r][c] = 2;return dfs(grid,r-1,c)+dfs(grid,r+1,c)+dfs(grid,r,c-1)+dfs(grid,r,c+1);}//判断[r,c]是否在网络中boolean inArea(int[][] grid,int r,int c){return 0<=r&&r<grid.length&&0<=c&&c<grid[0].length;}
}

695. 岛屿的最大面积 - 力扣（LeetCode）

class Solution {public int[] dx = {0,0,1,-1};public int[] dy = {1,-1,0,0};public int row,col;public boolean[][] check = new boolean[50][50];public int count = 0;public int ret = 0;public int maxAreaOfIsland(int[][] grid) {row = grid.length;col = grid[0].length;for(int i=0;i<row;i++){for(int j=0;j<col;j++){if(!check[i][j] && grid[i][j] ==1){count=0;//重置countdfs(grid,i,j);ret = Integer.max(ret,count);}}}return ret;}private void dfs(int[][] grid,int i,int j){count++;check[i][j] = true;for(int k=0;k<4;k++){int x = i+dx[k],y = j+dy[k];if(x>=0&&x<row&&y>=0&&y<col&&!check[x][y]&&grid[x][y]==1){dfs(grid,x,y);}}}
}

827. 最大人工岛 - 力扣（LeetCode）

class Solution {public int largestIsland(int[][] grid) {// dfs+哈希表//算法: 对所有岛屿编号,再遍历所有0,枚举所有能相连的岛屿的面积和最大值int n = grid.length;int maxArea = 0,islandIdx = 2;//1.对所有岛屿编号并记录在哈希表中Map<Integer,Integer>map = new HashMap<>();// key-岛屿编号 value-岛屿面积for(int i = 0;i<n;i++){for(int j = 0;j<n;j++){if(grid[i][j] == 1){int area = dfs(grid,i,j,islandIdx);map.put(islandIdx,area);islandIdx++;maxArea = Math.max(area,maxArea);}}}//2.枚举所有0的上下左右可能连接的情况for(int i = 0;i<n;i++){for(int j = 0;j<n;j++){Set<Integer>set = new HashSet<>();//记录/区分当前0周围的不同岛屿if(grid[i][j] == 0){//如果相邻的格子,没越界,且是岛屿,则把[岛屿编号]加入setif(i-1>=0&&grid[i-1][j] >= 2){set.add(grid[i-1][j]);//上}if(i+1<n&&grid[i+1][j]>=2){set.add(grid[i+1][j]);//下}if(j-1>=0&&grid[i][j-1]>=2){set.add(grid[i][j-1]);//左}if(j+1<n&&grid[i][j+1]>=2){set.add(grid[i][j+1]);//右}//初始化当前连接后的最大面积和int curMaxArea = 1;//for(int idx:set){int otherArea = map.get(idx);curMaxArea+=otherArea;}maxArea = Math.max(maxArea,curMaxArea);}}}return maxArea;}int dfs(int[][] grid,int x,int y,int islandIdx){//递归终止条件if(x<0||x>=grid.length||y<0||y>=grid[0].length||grid[x][y]!=1){return 0;}//标记已访问grid[x][y] = islandIdx;return 1+dfs(grid,x+1,y,islandIdx)+dfs(grid,x-1,y,islandIdx)+dfs(grid,x,y+1,islandIdx)+dfs(grid,x,y-1,islandIdx);}
}