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
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/number-of-islands
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
思路:深度优先搜索DFS
深度优先搜索 class Solution { public int numIslands(char[][] grid) { int res = 0; for(int i = 0; i < grid.length; i++) { for(int j = 0; j < grid[0].length; j ++) { if(grid[i][j] == '1') { dfs(grid, i, j); res ++; } } } return res; } public void dfs(char[][] grid, int i, int j) { if(i < 0 || i > grid.length - 1 || j < 0 || j > grid[0].length - 1 || grid[i][j] != '1') { return; } grid[i][j] = '2'; dfs(grid, i - 1, j); dfs(grid, i + 1, j); dfs(grid, i, j - 1); dfs(grid, i, j + 1); } }
类似题目:N皇后等