> 文档中心 > java深度优先搜索算法

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;     }}