代码随想录——图论一刷day02
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文章目录
- 前言
- 一、力扣695. 岛屿的最大面积
- 二、力扣1020. 飞地的数量
- 三、力扣1254. 统计封闭岛屿的数目
前言
`
一、力扣695. 岛屿的最大面积
淹没岛屿的递归
class Solution {
int[][] move = {
{0,1},
{0,-1},
{-1,0},
{1,0}
};
int count = 0;
public int maxAreaOfIsland(int[][] grid) {
int res = 0;
for(int i = 0; i < grid.length; i ++){
for(int j = 0; j < grid[i].length; j ++){
if(grid[i][j] == 1){
count = 1;
dfs(grid, i , j);
res = Math.max(count, res);
}
}
}
return res;
}
public void dfs(int[][] grid, int x, int y){
grid[x][y] = 0;
for(int i = 0; i < 4; i ++){
int nextX = x + move[i][0];
int nextY = y + move[i][1];
if(nextX < 0 || nextX >= grid.length || nextY < 0 || nextY >= grid[x].length || grid[nextX][nextY] == 0){
continue;
}
count ++;
dfs(grid, nextX, nextY);
}
}
}
使用标记数组的递归
class Solution {
int[][] move = {
{0,1},
{0,-1},
{-1,0},
{1,0}
};
boolean[][] flag;
int count = 0;
public int maxAreaOfIsland(int[][] grid) {
flag = new boolean[grid.length][grid[0].length];
int res = 0;
for(int i = 0; i < grid.length; i ++){
for(int j = 0; j < grid[i].length; j ++){
if(grid[i][j] == 1 && flag[i][j] == false){
count = 1;
dfs(grid, i , j);
res = Math.max(count, res);
}
}
}
return res;
}
public void dfs(int[][] grid, int x, int y){
flag[x][y] = true;
for(int i = 0; i < 4; i ++){
int nextX = x + move[i][0];
int nextY = y + move[i][1];
if(nextX < 0 || nextX >= grid.length || nextY < 0 || nextY >= grid[x].length || grid[nextX][nextY] == 0 || flag[nextX][nextY] == true){
continue;
}
count ++;
dfs(grid, nextX, nextY);
}
}
}
广度优先搜索
class Solution {
int[][] move = {
{0,1},
{0,-1},
{-1,0},
{1,0}
};
int count = 0;
boolean[][] flag;
public int maxAreaOfIsland(int[][] grid) {
int res = 0;
flag = new boolean[grid.length][grid[0].length];
for(int i = 0; i < grid.length; i ++){
for(int j = 0; j < grid[i].length; j ++){
if(grid[i][j] == 1 && flag[i][j] == false){
count = 1;
bfs(grid, i, j);
res = Math.max(res, count);
}
}
}
return res;
}
public void bfs(int[][] grid, int x, int y){
Deque<int[]> deq = new LinkedList<>();
flag[x][y] = true;
deq.offerLast(new int[]{x, y});
while(!deq.isEmpty()){
int[] cur = deq.pollFirst();
for(int i = 0; i < 4; i ++){
int nextX = cur[0] + move[i][0];
int nextY = cur[1] + move[i][1];
if(nextX < 0 || nextX >= grid.length || nextY < 0 || nextY >= grid[x].length || grid[nextX][nextY] == 0 || flag[nextX][nextY] == true){
continue;
}
count ++;
flag[nextX][nextY] = true;
deq.offerLast(new int[]{nextX, nextY});
}
}
}
}
二、力扣1020. 飞地的数量
DFS淹没岛屿同时一边统计边界岛屿的陆地面积， 一边统计总陆地面积，最后返回差值
class Solution {
int count = 0;
int[][] move = {
{0,1},
{0,-1},
{-1,0},
{1,0}
};
boolean tag = false;
public int numEnclaves(int[][] grid) {
int res = 0;
int edge = 0;
for(int i = 0; i < grid.length; i ++){
for(int j = 0; j < grid[i].length; j ++){
if(grid[i][j] == 1){
count = 1;
dfs(grid, i, j);
if(tag == true){
edge += count;
tag = false;
}
res += count;
}
}
}
return res - edge;
}
public void dfs(int[][] grid, int x, int y){
grid[x][y] = 0;
for(int i = 0; i < 4; i ++){
int nextX = move[i][0] + x;
int nextY = move[i][1] + y;
if(nextX < 0 || nextX >= grid.length || nextY < 0 || nextY >= grid[x].length){
tag = true;
}
if(nextX < 0 || nextX >= grid.length || nextY < 0 || nextY >= grid[x].length || grid[nextX][nextY] == 0){
continue;
}
count ++;
dfs(grid, nextX, nextY);
}
}
}
DFS使用标记数组同时一边统计边界岛屿的陆地面积， 一边统计总陆地面积，最后返回差值
class Solution {
int count = 0;
int[][] move = {
{0,1},
{0,-1},
{-1,0},
{1,0}
};
boolean[][] flag;
boolean tag = false;
public int numEnclaves(int[][] grid) {
flag = new boolean[grid.length][grid[0].length];
int res = 0;
int edge = 0;
for(int i = 0; i < grid.length; i ++){
for(int j = 0; j < grid[i].length; j ++){
if(grid[i][j] == 1 && flag[i][j] == false){
count = 1;
dfs(grid, i, j);
if(tag == true){
edge += count;
tag = false;
}
res += count;
}
}
}
return res - edge;
}
public void dfs(int[][] grid, int x, int y){
flag[x][y] = true;
for(int i = 0; i < 4; i ++){
int nextX = move[i][0] + x;
int nextY = move[i][1] + y;
if(nextX < 0 || nextX >= grid.length || nextY < 0 || nextY >= grid[x].length){
tag = true;
}
if(nextX < 0 || nextX >= grid.length || nextY < 0 || nextY >= grid[x].length || grid[nextX][nextY] == 0 || flag[nextX][nextY] == true){
continue;
}
count ++;
dfs(grid, nextX, nextY);
}
}
}
BFS使用标记数组同时一边统计边界岛屿的陆地面积， 一边统计总陆地面积，最后返回差值
class Solution {
int count = 0;
int[][] move = {
{0,1},
{0,-1},
{-1,0},
{1,0}
};
boolean[][] flag;
boolean tag = false;
public int numEnclaves(int[][] grid) {
flag = new boolean[grid.length][grid[0].length];
int res = 0;
int edge = 0;
for(int i = 0; i < grid.length; i ++){
for(int j = 0; j < grid[i].length; j ++){
if(grid[i][j] == 1 && flag[i][j] == false){
count = 1;
bfs(grid, i, j);
if(tag == true){
edge += count;
tag = false;
}
res += count;
}
}
}
return res - edge;
}
public void bfs(int[][] grid, int x, int y){
flag[x][y] = true;
Deque<int[]> deq = new LinkedList<>();
deq.offerLast(new int[]{x,y});
while(!deq.isEmpty()){
int[] cur = deq.pollFirst();
for(int i = 0; i < 4; i ++){
int nextX = move[i][0] + cur[0];
int nextY = move[i][1] + cur[1];
if(nextX < 0 || nextX >= grid.length || nextY < 0 || nextY >= grid[x].length){
tag = true;
}
if(nextX < 0 || nextX >= grid.length || nextY < 0 || nextY >= grid[x].length || grid[nextX][nextY] == 0 || flag[nextX][nextY] == true){
continue;
}
count ++;
flag[nextX][nextY] = true;
deq.offerLast(new int[]{nextX,nextY});
}
}
}
}
三、力扣1254. 统计封闭岛屿的数目
class Solution {
int[][] move = {
{1,0},
{-1,0},
{0,1},
{0,-1}
};
public int closedIsland(int[][] grid) {
int res = 0;
for(int i = 0; i < grid.length; i ++){
if(grid[i][0] == 0){
dfs(grid, i, 0);
}
if(grid[i][grid[0].length-1] == 0){
dfs(grid, i, grid[0].length-1);
}
}
for(int i = 0; i < grid[0].length; i ++){
if(grid[0][i] == 0){
dfs(grid, 0, i);
}
if(grid[grid.length-1][i] == 0){
dfs(grid, grid.length-1, i);
}
}
for(int i = 0; i < grid.length; i ++){
for(int j = 0; j < grid[0].length; j ++){
if(grid[i][j] == 0){
res ++;
dfs(grid, i, j);
}
}
}
return res;
}
public void dfs(int[][] grid, int x, int y){
grid[x][y] = 1;
for(int i = 0; i < 4; i ++){
int nextX = move[i][0] + x;
int nextY = move[i][1] + y;
if(nextX < 0 || nextX >= grid.length || nextY < 0 || nextY >= grid[x].length || grid[nextX][nextY] == 1){
continue;
}
dfs(grid, nextX, nextY);
}
}
}