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文章目录
- 前言
- [一、力扣200. 岛屿数量](#一、力扣200. 岛屿数量)
- [二、力扣695. 岛屿的最大面积](#二、力扣695. 岛屿的最大面积)
- [三、力扣1020. 飞地的数量](#三、力扣1020. 飞地的数量)
- [四、力扣130. 被围绕的区域](#四、力扣130. 被围绕的区域)
前言
依然是从地图周边出发,将周边空格相邻的'O'都做上标记,然后在遍历一遍地图,遇到 'O' 且没做过标记的,那么都是地图中间的'O',全部改成'X'就行。
一、力扣200. 岛屿数量
java
class Solution {
int[][] arr = new int[][]{{0,1},{0,-1},{-1,0},{1,0}};
boolean[][] flag;
public int numIslands(char[][] grid) {
int m = grid.length, n = grid[0].length;
int res = 0;
flag = new boolean[m][n];
for(int i = 0; i < m; i ++){
for(int j = 0; j < n; j ++){
if(!flag[i][j] && grid[i][j] == '1'){
res ++;
bfs(grid, i, j);
}
}
}
return res;
}
public void bfs(char[][] grid, int x, int y){
Deque<int[]> deq = new LinkedList<>();
deq.offerLast(new int[]{x,y});
while(!deq.isEmpty()){
int size = deq.size();
for(int i = 0; i < size; i ++){
int[] cur = deq.pollFirst();
for(int j = 0; j < 4; j ++){
int curX = cur[0] + arr[j][0];
int curY = cur[1] + arr[j][1];
if(curX < 0 || curX >= grid.length || curY < 0 || curY >= grid[0].length){
continue;
}
if(!flag[curX][curY] && grid[curX][curY] == '1'){
flag[curX][curY] = true;
deq.offerLast(new int[]{curX,curY});
}
}
}
}
}
}
二、力扣695. 岛屿的最大面积
java
class Solution {
int res = 0;
int count = 0;
int[][] arr = new int[][]{{0,1},{0,-1},{-1,0},{1,0}};
boolean[][] flag;
public int maxAreaOfIsland(int[][] grid) {
int m = grid.length, n = grid[0].length;
flag = new boolean[m][n];
for(int i = 0; i < m; i ++){
for(int j = 0; j < n; j ++){
if(!flag[i][j] && grid[i][j] == 1){
flag[i][j] = true;
count = 1;
dfs(grid,i,j);
}
}
}
return res;
}
public void dfs(int[][] grid, int x, int y){
res = Math.max(count,res);
for(int i = 0; i < 4; i ++){
int curX = x + arr[i][0];
int curY = y + arr[i][1];
if(curX < 0 || curX >= grid.length || curY < 0 || curY >= grid[0].length){
continue;
}
if(!flag[curX][curY] && grid[curX][curY] == 1){
flag[curX][curY] = true;
count ++;
dfs(grid,curX,curY);
}
}
}
}
三、力扣1020. 飞地的数量
java
class Solution {
int res = 0;
int count = 0;
int[][] arr = new int[][]{{0,1},{0,-1},{-1,0},{1,0}};
boolean[][] flag;
boolean f;
public int numEnclaves(int[][] grid) {
int m = grid.length, n = grid[0].length;
flag = new boolean[m][n];
for(int i = 0; i < m; i ++){
for(int j = 0; j < n; j ++){
if(!flag[i][j] && grid[i][j] == 1){
flag[i][j] = true;
count = 1;
f = false;
dfs(grid,i,j);
if(!f){
res += count;
}
}
}
}
return res;
}
public void dfs(int[][] grid, int x, int y){
for(int i = 0; i < 4; i ++){
int curX = x + arr[i][0];
int curY = y + arr[i][1];
if(curX < 0 || curX >= grid.length || curY < 0 || curY >= grid[0].length){
f = true;
continue;
}
if(!flag[curX][curY] && grid[curX][curY] == 1){
count ++;
flag[curX][curY] = true;
dfs(grid,curX,curY);
}
}
}
}
四、力扣130. 被围绕的区域
java
class Solution {
int[][] arr = new int[][]{{0,1},{0,-1},{-1,0},{1,0}};
boolean[][] flag;
public void solve(char[][] board) {
int m = board.length, n = board[0].length;
flag = new boolean[m][n];
for(int i = 0; i < m; i ++){
if(!flag[i][0] && board[i][0] == 'O'){
flag[i][0] = true;
dfs(board, i, 0);
}
if(!flag[i][n-1] && board[i][n-1] == 'O'){
flag[i][n-1] = true;
dfs(board,i,n-1);
}
}
for(int i = 0; i < n; i ++){
if(!flag[0][i] && board[0][i] == 'O'){
flag[0][i] = true;
dfs(board,0,i);
}
if(!flag[m-1][i] && board[m-1][i] == 'O'){
flag[m-1][i] = true;
dfs(board,m-1,i);
}
}
for(int i = 0; i < m; i ++){
for(int j = 0; j < n; j ++){
if(!flag[i][j] && board[i][j] == 'O'){
board[i][j] = 'X';
}
}
}
}
public void dfs(char[][] board, int x, int y){
for(int i = 0; i < 4; i++){
int curX = x + arr[i][0];
int curY = y + arr[i][1];
if(curX < 0 || curX >= board.length || curY < 0 || curY >= board[0].length){
continue;
}
if(!flag[curX][curY] && board[curX][curY] == 'O'){
flag[curX][curY] = true;
dfs(board,curX,curY);
}
}
}
}