更新 0103.水流问题 DFS解法Java版本

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2024-07-17 22:31:23 -07:00
parent b80f731427
commit 0ed1eb394b

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