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Given a 2d grid map of ‘1‘s (land) and ‘0‘s (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.
Example 1:
11110
11010
11000
00000
Answer: 1
Example 2:
11000
11000
00100
00011
Answer: 3
哈哈，这是今天的新题，十多分钟A了，题目不难，我的做法是BFS+DFS，题目大意是1代表岛屿，0代表水，可以假设这个grid周围都是水，岛屿定义是四周都是水，求岛屿的个数。
我的做法是，从第一个开始，如果是1，并且没有访问过visit为false，那么加入queue里，当queue非空取出来并加入它上下左右的邻居，并且把这些节点visit设置为true，时间复杂度是O(M*N)，空间复杂度的话，因为同时在queue里的最多也就4个节点的坐标，所以是O(1)。
Talk is cheap>>
public int numIslands(char[][] grid) { if (grid == null || grid.length == 0) { return 0; } int rowLen = grid.length; int colLen = grid[0].length; boolean[][] visited = new boolean[rowLen][colLen]; int[][] direct = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}}; Deque<Integer> queue = new ArrayDeque<>(); int res = 0; for (int i = 0; i < rowLen; i++) { for (int j = 0; j < colLen; j++) { if (!visited[i][j] && grid[i][j] == ‘1‘) { res++; queue.add(i * colLen + j); while (!queue.isEmpty()) { int pos = queue.poll(); int curr_x = pos / colLen; int curr_y = pos % colLen; if (visited[curr_x][curr_y]) { continue; } visited[curr_x][curr_y] = true; for (int k = 0; k < 4; k++) { int x = curr_x + direct[k][0]; int y = curr_y + direct[k][1]; if (x >= 0 && y >= 0 && x < rowLen && y < colLen && grid[x][y] == ‘1‘) { queue.add(x * colLen + y); } } } } } } return res; }
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原文地址：http://www.cnblogs.com/aboutblank/p/4405557.html
