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更新 0103.水流问题 DFS解法Java版本
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@ -281,6 +281,78 @@ for (int j = 0; j < m; j++) {
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## 其他语言版本
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### Java
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```Java
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public class Main {
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// 采用 DFS 进行搜索
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public static void dfs(int[][] heights, int x, int y, boolean[][] visited, int preH) {
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// 遇到边界或者访问过的点,直接返回
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if (x < 0 || x >= heights.length || y < 0 || y >= heights[0].length || visited[x][y]) return;
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// 不满足水流入条件的直接返回
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if (heights[x][y] < preH) return;
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// 满足条件,设置为true,表示可以从边界到达此位置
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visited[x][y] = true;
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// 向下一层继续搜索
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dfs(heights, x + 1, y, visited, heights[x][y]);
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dfs(heights, x - 1, y, visited, heights[x][y]);
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dfs(heights, x, y + 1, visited, heights[x][y]);
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dfs(heights, x, y - 1, visited, heights[x][y]);
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}
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public static void main(String[] args) {
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Scanner sc = new Scanner(System.in);
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int m = sc.nextInt();
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int n = sc.nextInt();
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int[][] heights = new int[m][n];
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for (int i = 0; i < m; i++) {
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for (int j = 0; j < n; j++) {
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heights[i][j] = sc.nextInt();
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}
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}
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// 初始化两个二位boolean数组,代表两个边界
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boolean[][] pacific = new boolean[m][n];
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boolean[][] atlantic = new boolean[m][n];
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// 从左右边界出发进行DFS
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for (int i = 0; i < m; i++) {
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dfs(heights, i, 0, pacific, Integer.MIN_VALUE);
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dfs(heights, i, n - 1, atlantic, Integer.MIN_VALUE);
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}
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// 从上下边界出发进行DFS
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for (int j = 0; j < n; j++) {
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dfs(heights, 0, j, pacific, Integer.MIN_VALUE);
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dfs(heights, m - 1, j, atlantic, Integer.MIN_VALUE);
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}
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// 当两个边界二维数组在某个位置都为true时,符合题目要求
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List<List<Integer>> res = new ArrayList<>();
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for (int i = 0; i < m; i++) {
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for (int j = 0; j < n; j++) {
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if (pacific[i][j] && atlantic[i][j]) {
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res.add(Arrays.asList(i, j));
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}
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}
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}
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// 打印结果
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for (List<Integer> list : res) {
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for (int k = 0; k < list.size(); k++) {
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if (k == 0) {
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System.out.print(list.get(k) + " ");
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} else {
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System.out.print(list.get(k));
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}
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}
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System.out.println();
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}
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}
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}
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```
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### Python
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