【問題描述】[中等]

【解答思路】

1. 深度優(yōu)先搜索

使用深度優(yōu)先搜索實現(xiàn)標(biāo)記操作。在下面的代碼中，我們把標(biāo)記過的字母 O 修改為字母 A。
復(fù)雜度

class Solution {int[] dx = {1, -1, 0, 0};int[] dy = {0, 0, 1, -1};public void solve(char[][] board) {int n = board.length;if (n == 0) {return;}int m = board[0].length;Queue<int[]> queue = new LinkedList<int[]>();for (int i = 0; i < n; i++) {if (board[i][0] == 'O') {queue.offer(new int[]{i, 0});}if (board[i][m - 1] == 'O') {queue.offer(new int[]{i, m - 1});}}for (int i = 1; i < m - 1; i++) {if (board[0][i] == 'O') {queue.offer(new int[]{0, i});}if (board[n - 1][i] == 'O') {queue.offer(new int[]{n - 1, i});}}while (!queue.isEmpty()) {int[] cell = queue.poll();int x = cell[0], y = cell[1];board[x][y] = 'A';for (int i = 0; i < 4; i++) {int mx = x + dx[i], my = y + dy[i];if (mx < 0 || my < 0 || mx >= n || my >= m || board[mx][my] != 'O') {continue;}queue.offer(new int[]{mx, my});}}for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {if (board[i][j] == 'A') {board[i][j] = 'O';} else if (board[i][j] == 'O') {board[i][j] = 'X';}}}} }

2. 廣度優(yōu)先搜索

以使用廣度優(yōu)先搜索實現(xiàn)標(biāo)記操作。在下面的代碼中，我們把標(biāo)記過的字母 O 修改為字母 A
時間復(fù)雜度：O(N) 空間復(fù)雜度：O(1)

class Solution {int[] dx = {1, -1, 0, 0};int[] dy = {0, 0, 1, -1};public void solve(char[][] board) {int n = board.length;if (n == 0) {return;}int m = board[0].length;Queue<int[]> queue = new LinkedList<int[]>();for (int i = 0; i < n; i++) {if (board[i][0] == 'O') {queue.offer(new int[]{i, 0});}if (board[i][m - 1] == 'O') {queue.offer(new int[]{i, m - 1});}}for (int i = 1; i < m - 1; i++) {if (board[0][i] == 'O') {queue.offer(new int[]{0, i});}if (board[n - 1][i] == 'O') {queue.offer(new int[]{n - 1, i});}}while (!queue.isEmpty()) {int[] cell = queue.poll();int x = cell[0], y = cell[1];board[x][y] = 'A';for (int i = 0; i < 4; i++) {int mx = x + dx[i], my = y + dy[i];if (mx < 0 || my < 0 || mx >= n || my >= m || board[mx][my] != 'O') {continue;}queue.offer(new int[]{mx, my});}}for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {if (board[i][j] == 'A') {board[i][j] = 'O';} else if (board[i][j] == 'O') {board[i][j] = 'X';}}}} }

【總結(jié)】

1. 細(xì)節(jié)：

1.1 方向定義
int[] dx = {1, -1, 0, 0};
int[] dy = {0, 0, 1, -1};
1.2 邊界判斷
if (mx < 0 || my < 0 || mx >= n || my >= m || board[mx][my] != ‘O’) {
continue;
}

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2.DFS BFS 思路不復(fù)雜 注意細(xì)節(jié) 多寫幾遍

轉(zhuǎn)載鏈接：https://leetcode-cn.com/problems/surrounded-regions/solution/bei-wei-rao-de-qu-yu-by-leetcode-solution/

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