package top100.图论;import java.util.LinkedList;
import java.util.Queue;public class TOP {//200. 岛屿数量//网格类问题的 DFS 遍历方法int[][] ways = new int[][]{{-1, 0}, {1, 0}, {0, 1}, {0, -1}};public int numIslands(char[][] grid) {int m = grid.length, n = grid[0].length;int res = 0;for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {if (grid[i][j] == '1') {res++;dfs(grid, i, j, m, n);}}}return res;}private void dfs(char[][] grid, int i, int j, int m, int n) {grid[i][j] = '0';for (int[] way : ways) {int x = i + way[0], y = j + way[1];if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == '1') {dfs(grid, x, y, m, n);}}}// 994. 腐烂的橘子// BFS遍历图//同时遍历所有的腐烂橘子，类似对k个有序链表排序问题的处理思想，用队列记录当前要处理的结点public int orangesRotting(int[][] grid) {int m = grid.length;int n = grid[0].length;int fresh = 0; //统计新鲜的橘子个数Queue<int[]> queue = new LinkedList<>(); //记录每轮腐烂的橘子位置for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {if (grid[i][j] == 1) {fresh++;} else if (grid[i][j] == 2) {queue.add(new int[]{i, j});}}}int[][] ways = new int[][]{{1, 0}, {-1, 0}, {0, 1}, {0, -1}};int res = 0;while (fresh > 0 && !queue.isEmpty()) { //while (fresh > 0)会有死循环,有无法感染到的橘子res++;int k = queue.size();//每轮取出上一轮感染的橘子for (int i = 0; i < k; i++) {int[] cur = queue.poll();for (int[] way : ways) {int x = cur[0] + way[0];int y = cur[1] + way[1];//可以感染的新鲜橘子if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1) {fresh--;grid[x][y] = 2;//将本轮感染的橘子标记，防止重复遍历queue.add(new int[]{x, y});}}}}//有橘子无法感染if (fresh > 0) {return -1;} else {return res;}}
}