java深度优先搜索算法
题目1
给你一个由 '1'(陆地)和 '0'(水)组成的的二维网格,请你计算网格中岛屿的数量。
岛屿总是被水包围,并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。
此外,你可以假设该网格的四条边均被水包围。
示例 1:
输入:grid = [
["1","1","1","1","0"],
["1","1","0","1","0"],
["1","1","0","0","0"],
["0","0","0","0","0"]
]
输出:1
示例 2:
输入:grid = [
["1","1","0","0","0"],
["1","1","0","0","0"],
["0","0","1","0","0"],
["0","0","0","1","1"]
]
输出:3
思想:先遍历找到一个一个1,通过dfs不断将这个1上下左右的1变成0,这样它们就合并成一个岛屿。
class Solution { public int numIslands(char[][] grid) { if(grid == null || grid.length == 0) return 0; int len1 = grid.length; int len2 = grid[0].length; int count = 0; for(int i = 0;i < len1;i++) { for(int j = 0;j < len2;j++) { if(grid[i][j] == '1') { count++; dfs(grid,i,j); } } } return count; } void dfs(char[][] grid,int i,int j) { int len1 = grid.length; int len2 = grid[0].length; if(i < 0 || j = len1 || j >= len2 || grid[i][j] == '0') return; grid[i][j] = '0'; dfs(grid,i+1,j); dfs(grid,i-1,j); dfs(grid,i,j+1); dfs(grid,i,j-1); }}
题目:
给定一个 row x col 的二维网格地图 grid ,其中:grid[i][j] = 1 表示陆地, grid[i][j] = 0 表示水域。
网格中的格子 水平和垂直 方向相连(对角线方向不相连)。整个网格被水完全包围,但其中恰好有一个岛屿(或者说,一个或多个表示陆地的格子相连组成的岛屿)。
岛屿中没有“湖”(“湖” 指水域在岛屿内部且不和岛屿周围的水相连)。格子是边长为 1 的正方形。网格为长方形,且宽度和高度均不超过 100 。计算这个岛屿的周长。
class dfs { static int[] dx = {0, 1, 0, -1}; static int[] dy = {1, 0, -1, 0}; public int islandPerimeter(int[][] grid) { int n = grid.length, m = grid[0].length; int ans = 0; for (int i = 0; i < n; ++i) { for (int j = 0; j < m; ++j) { if (grid[i][j] == 1) { ans += dfs(i, j, grid, n, m); } } } return ans; } public int dfs(int x, int y, int[][] grid, int n, int m) { //边界+1 if (x = n || y = m || grid[x][y] == 0) { return 1; } //已经遍历过的 if (grid[x][y] == 2) { return 0; } grid[x][y] = 2; int res = 0; //四周开始遍历 for (int i = 0; i < 4; ++i) { int tx = x + dx[i]; int ty = y + dy[i]; res += dfs(tx, ty, grid, n, m); } return res; }}