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leetcode 200 岛屿数量

幸福的无所谓 2022-04-16 阅读 52
java

leetcode 200 岛屿数量
思路:深度优先搜索,遍历整个矩阵,遍历到的值如果是1并且不在已访问列表中,岛屿数量+1,则对这个值的四周进行深度优先搜索,搜索到的四周的值如果是1则加入已访问列表。

class Solution {
    public void dfs(boolean[][] visited, char[][] grid, int i, int j){
        int row = grid.length;
        int column = grid[0].length;

        visited[i][j]=true;
        //如果不写!visited[i-1][j]&& grid[i-1][j]=='1'则出现stackOverflow错误,因为循环调用
        
        if(0<=i-1 && i-1<row && !visited[i-1][j] && grid[i-1][j]=='1'){
            dfs(visited,grid,i-1,j);
        }
        if(0<=i+1 && i+1<row && !visited[i+1][j] && grid[i+1][j]=='1'){
            dfs(visited,grid,i+1,j);
        }
        if(0<=j-1 && j-1<column && !visited[i][j-1] && grid[i][j-1]=='1'){
            dfs(visited,grid,i,j-1);
        }
        if(0<=j+1 && j+1<column && !visited[i][j+1] && grid[i][j+1]=='1'){
            dfs(visited,grid,i,j+1);
        }

    }
    public int numIslands(char[][] grid) {
        int island = 0;
        int row = grid.length;
        int column = grid[0].length;
        boolean[][] visited = new boolean[row][column];
        for(int i = 0; i < row; i ++){
            for(int j = 0; j < column; j++){
                if(grid[i][j]=='1' && !visited[i][j]){

                    island++;
                    dfs(visited,grid,i,j);
                }
            }
        }
        return island;
    }
}
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