104 lines
3.4 KiB
Java
104 lines
3.4 KiB
Java
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//给你一个 m x n 的矩阵 board ，由若干字符 'X' 和 'O' ，找到所有被 'X' 围绕的区域，并将这些区域里所有的 'O' 用 'X' 填充
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//。
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//
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//
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//
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//
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// 示例 1：
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//
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//
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//输入：board = [["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X"
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//,"X"]]
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//输出：[["X","X","X","X"],["X","X","X","X"],["X","X","X","X"],["X","O","X","X"]]
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//解释：被围绕的区间不会存在于边界上，换句话说，任何边界上的 'O' 都不会被填充为 'X'。 任何不在边界上，或不与边界上的 'O' 相连的 'O' 最终都
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//会被填充为 'X'。如果两个元素在水平或垂直方向相邻，则称它们是“相连”的。
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//
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//
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// 示例 2：
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//
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//
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//输入：board = [["X"]]
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//输出：[["X"]]
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//
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//
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//
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//
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// 提示：
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//
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//
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// m == board.length
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// n == board[i].length
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// 1 <= m, n <= 200
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// board[i][j] 为 'X' 或 'O'
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//
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//
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//
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// Related Topics 深度优先搜索 广度优先搜索 并查集 数组 矩阵
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// 👍 587 👎 0
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package leetcode.editor.cn;
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import javafx.util.Pair;
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import java.util.LinkedList;
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import java.util.Queue;
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//130:被围绕的区域
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class SurroundedRegions {
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public static void main(String[] args) {
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//测试代码
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Solution solution = new SurroundedRegions().new Solution();
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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class Solution {
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public void solve(char[][] board) {
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boolean[][] use = new boolean[board.length][board[0].length];
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Queue<Pair<Integer, Integer>> queue = new LinkedList<>();
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for (int i = 0; i < board.length; i++) {
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if (board[i][0] == 'O') {
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queue.add(new Pair<>(i, 0));
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use[i][0] = true;
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}
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if (board[i][board[0].length - 1] == 'O') {
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queue.add(new Pair<>(i, board[0].length - 1));
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use[i][board[0].length - 1] = true;
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}
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}
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for (int i = 1; i < board[0].length - 1; i++) {
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if (board[0][i] == 'O') {
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queue.add(new Pair<>(0, i));
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use[0][i] = true;
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}
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if (board[board.length - 1][i] == 'O') {
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queue.add(new Pair<>(board.length - 1, i));
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use[board.length - 1][i] = true;
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}
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}
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int[] xIndex = new int[]{-1, 1, 0, 0};
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int[] yIndex = new int[]{0, 0, -1, 1};
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while (!queue.isEmpty()) {
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Pair<Integer, Integer> pair = queue.poll();
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for (int i = 0; i < 4; i++) {
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int x = pair.getKey() + xIndex[i];
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int y = pair.getValue() + yIndex[i];
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if (x < 0 || x >= board.length || y < 0 || y >= board[0].length || use[x][y] || board[x][y] == 'X') {
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continue;
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}
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queue.add(new Pair<>(x, y));
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use[x][y] = true;
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}
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}
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for (int i = 0; i < board.length; i++) {
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for (int j = 0; j < board[0].length; j++) {
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if (board[i][j] == 'O' && !use[i][j]) {
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board[i][j] = 'X';
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}
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}
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}
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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}
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