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添加1020.飞抵的数量Java版本的代码
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@ -144,6 +144,130 @@ public:
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}
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};
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```
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## 其他语言版本
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**Java**:
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深度优先遍历版本:
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```java
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class Solution {
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// 四个方向
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private static final int[][] position = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
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// 深度优先遍历,把可以通向边缘部分的 1 全部标记成 true
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public void dfs(int[][] grid, int row, int col, boolean[][] visited) {
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for (int[] current: position) {
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int newRow = row + current[0], newCol = col + current[1];
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// 下标越界直接跳过
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if (newRow < 0 || newRow >= grid.length || newCol < 0 || newCol >= grid[0].length) continue;
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// 当前位置不是 1 或者已经被访问了就直接跳过
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if (grid[newRow][newCol] != 1 || visited[newRow][newCol]) continue;
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visited[newRow][newCol] = true;
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dfs(grid, newRow, newCol, visited);
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}
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}
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public int numEnclaves(int[][] grid) {
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int rowSize = grid.length, colSize = grid[0].length, ans = 0; // ans 记录答案
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boolean[][] visited = new boolean[rowSize][colSize]; // 标记数组
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// 左侧边界和右侧边界查找 1 进行标记并进行深度优先遍历
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for (int row = 0; row < rowSize; row++) {
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if (grid[row][0] == 1 && !visited[row][0]) {
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visited[row][0] = true;
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dfs(grid, row, 0, visited);
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}
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if (grid[row][colSize - 1] == 1 && !visited[row][colSize - 1]) {
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visited[row][colSize - 1] = true;
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dfs(grid, row, colSize - 1, visited);
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}
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}
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// 上边界和下边界遍历,但是四个角不用遍历,因为上面已经遍历到了
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for (int col = 1; col < colSize - 1; col++) {
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if (grid[0][col] == 1 && !visited[0][col]) {
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visited[0][col] = true;
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dfs(grid, 0, col, visited);
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}
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if (grid[rowSize - 1][col] == 1 && !visited[rowSize - 1][col]) {
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visited[rowSize - 1][col] = true;
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dfs(grid, rowSize - 1, col, visited);
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}
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}
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// 查找没有标记过的 1,记录到 ans 中
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for (int row = 0; row < rowSize; row++) {
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for (int col = 0; col < colSize; col++) {
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if (grid[row][col] == 1 && !visited[row][col]) ++ans;
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}
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}
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return ans;
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}
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}
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```
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广度优先遍历版本:
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```java
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class Solution {
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// 四个方向
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private static final int[][] position = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
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// 广度优先遍历,把可以通向边缘部分的 1 全部标记成 true
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public void bfs(int[][] grid, Queue<int[]> queue, boolean[][] visited) {
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while (!queue.isEmpty()) {
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int[] curPos = queue.poll();
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for (int[] current: position) {
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int row = curPos[0] + current[0], col = curPos[1] + current[1];
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// 下标越界直接跳过
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if (row < 0 || row >= grid.length || col < 0 || col >= grid[0].length)
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continue;
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// 当前位置不是 1 或者已经被访问了就直接跳过
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if (visited[row][col] || grid[row][col] == 0) continue;
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visited[row][col] = true;
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queue.add(new int[]{row, col});
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}
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}
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}
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public int numEnclaves(int[][] grid) {
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int rowSize = grid.length, colSize = grid[0].length, ans = 0; // ans 记录答案
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boolean[][] visited = new boolean[rowSize][colSize]; // 标记数组
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Queue<int[]> queue = new ArrayDeque<>();
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// 左侧边界和右侧边界查找 1 进行标记并进行深度优先遍历
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for (int row = 0; row < rowSize; row++) {
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if (grid[row][0] == 1) {
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visited[row][0] = true;
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queue.add(new int[]{row, 0});
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}
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if (grid[row][colSize - 1] == 1) {
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visited[row][colSize - 1] = true;
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queue.add(new int[]{row, colSize - 1});
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}
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}
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// 上边界和下边界遍历,但是四个角不用遍历,因为上面已经遍历到了
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for (int col = 1; col < colSize - 1; col++) {
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if (grid[0][col] == 1) {
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visited[0][col] = true;
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queue.add(new int[]{0, col});
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}
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if (grid[rowSize - 1][col] == 1 && !visited[rowSize - 1][col]) {
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visited[rowSize - 1][col] = true;
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queue.add(new int[]{rowSize - 1, col});
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}
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}
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bfs(grid, queue, visited);
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// 查找没有标记过的 1,记录到 ans 中
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for (int row = 0; row < rowSize; row++) {
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for (int col = 0; col < colSize; col++) {
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if (grid[row][col] == 1 && !visited[row][col]) ++ans;
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}
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}
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return ans;
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}
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}
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```
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## 类似题目
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* 1254. 统计封闭岛屿的数目
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@ -153,3 +277,4 @@ public:
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<a href="https://programmercarl.com/other/kstar.html" target="_blank">
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<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
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</a>
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