力扣-695-岛屿的最大面积
import org.junit.Test; import java.util.*; public class leetcode695 { int[] par = new int[100005]; int[] rak = new int[100005]; int[] dx = {0,1,0,-1}; int[] dy = {1,0,-1,0}; // 并查集的初始化 public void init(int n) { for (int i = 0; i < n; i++) { par[i] = i; rak[i] = 0; } } // 并查集的根节点查询 public int find(int x) { if (par[x] == x) return x; return par[x] = find(par[x]); } // 并查集的合并 public void uinte(int x, int y) { x = find(x); y = find(y); if (x == y) return; if (rak[x] < rak[y]) { par[x] = y; } else { par[y] = x; if (rak[x] == rak[y]) rak[x]++; } } // 并查集的判断 public boolean same(int x, int y) { return find(x) == find(y); } public int maxAreaOfIsland(int[][] grid) { if (grid.length == 0 || grid[0].length == 0) return 0; int row = grid.length; int col = grid[0].length; init(row*col); // 创建连通块 for (int i = 0; i < row; i++) { for (int j = 0; j < col; j++) { if (grid[i][j] == 1) { for (int k = 0; k < 4; k++) { int xx = i + dx[k]; int yy = j + dy[k]; if (isValid(xx, yy, row, col) && grid[xx][yy] == 1 && !same(i*col+j, xx*col+yy)) { uinte(i*col+j, xx*col+yy); } } } } } // HashSet保存每个连通块的root HashSet<Integer> set = new HashSet<>(); for (int i = 0; i < row*col; i++) { if (par[i] == i && grid[i/col][i%col] == 1) set.add(i); } // 如果没有岛屿 if (set.isEmpty()) return 0; System.out.println(set); // 保存每个连通块的id及其面积 HashMap<Integer, Integer> map = new HashMap<>(); for (int root: set) { int num = 0; for (int i = 0; i < row; i++) { for (int j = 0; j < col; j++) { // System.out.println(par[i*col+j]); if (find(i*col+j) == root) num++; } } map.put(root, num); } // 寻找面积最大的岛屿 int res = 0; for (Map.Entry<Integer, Integer> entry: map.entrySet()) { res = Math.max(res, entry.getValue()); } return res; } // 坐标合法性检验 public boolean isValid(int x, int y, int row, int col) { if (x < 0 || x >= row || y < 0 || y >= col) return false; return true; } @Test public void test() { int[][] grid = {{1,1,1}, {1,1,1}, {1,1,1}}; System.out.println(maxAreaOfIsland_bfs(grid)); } }
import org.junit.Test; import java.util.*; public class leetcode695 { @Test public void test() { int[][] grid = {{1,1,1}, {1,1,1}, {1,1,1}}; System.out.println(maxAreaOfIsland_bfs(grid)); } // 方法二、深度优先搜索 public int maxAreaOfIsland_dfs(int[][] grid) { if (grid.length == 0 || grid[0].length == 0) return 0; int row = grid.length; int col = grid[0].length; int res = 0; for (int i = 0; i < row; i++) { for (int j = 0; j < col; j++) { res = Math.max(res, dfs(grid, i, j)); } } return res; } public int dfs(int[][] grid, int x,int y) { if (x < 0 || x >= grid.length || y < 0 || y >= grid[0].length || grid[x][y] == 0) return 0; int[] dx = {0,1,0,-1}; int[] dy = {1,0,-1,0}; // 记录当前位置的岛屿数量 int res = grid[x][y]; // dfs遍历过的地方变为0 grid[x][y] = 0; for (int k = 0; k < 4; k++) { int xx = x + dx[k]; int yy = y + dy[k]; res += dfs(grid, xx, yy); } return res; } }
import org.junit.Test; import java.util.*; public class leetcode695 { // 方法三、广度优先搜索(基于队列实现) class coordinate { int x; int y; public coordinate(int x, int y) { this.x = x; this.y = y; } } public int maxAreaOfIsland_bfs(int[][] grid) { if (grid.length == 0 || grid[0].length == 0) return 0; int row = grid.length; int col = grid[0].length; int[] dx = {0,1,0,-1}; int[] dy = {1,0,-1,0}; int res = 0; for (int i = 0; i < row; i++) { for (int j = 0; j < col; j++) { if (grid[i][j] == 0) continue; int cur = 0; Queue<coordinate> que = new LinkedList<>(); que.offer(new coordinate(i,j)); cur += grid[i][j]; grid[i][j] = 0; while (!que.isEmpty()) { coordinate pos = que.poll(); for (int k = 0; k < 4; k++) { int xx = pos.x + dx[k]; int yy = pos.y + dy[k]; if (xx < 0 || xx >= row || yy < 0 || yy >= col || grid[xx][yy] == 0) continue; que.offer(new coordinate(xx, yy)); cur += grid[xx][yy]; grid[xx][yy]=0; } } res = Math.max(res, cur); } } return res; } }
作者:Ryanjie
出处:http://www.cnblogs.com/ryanjan/
本文版权归作者和博客园所有,欢迎转载。转载请在留言板处留言给我,且在文章标明原文链接,谢谢!
如果您觉得本篇博文对您有所收获,觉得我还算用心,请点击右下角的 [推荐],谢谢!