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  • 队列 & 栈//岛屿的个数

    给定一个由 '1'(陆地)和 '0'(水)组成的的二维网格,计算岛屿的数量。一个岛被水包围,并且它是通过水平方向或垂直方向上相邻的陆地连接而成的。你可以假设网格的四个边均被水包围。

    示例 1:

    输入:
    11110
    11010
    11000
    00000
    
    输出: 1
    

    示例 2:

    输入:
    11000
    11000
    00100
    00011
    
    输出: 3
    class Solution {
        public int numIslands(char[][] grid) {
            if(grid.length==0||grid[0].length==0) return 0;
            int m = grid.length;
            int n = grid[0].length;
            int res = 0;
            for(int i = 0; i < m; i++){
                for(int j = 0; j < n; j++){
                    if(grid[i][j] == '1')
                    {
                        dfs(grid,i,j);
                        res++;
                    }
                }
            }
            return res;
        }
        public void dfs(char[][] g, int x, int y){
            if(x<0||x>=g.length) return ;
            if(y<0||y>=g[0].length) return ;
            if(g[x][y]!='1')
                return ;
            g[x][y] = '0';
            dfs(g,x+1,y);
            dfs(g,x-1,y);
            dfs(g,x,y+1);
            dfs(g,x,y-1);
        }
    }
    class Solution {
    public:
        int numIslands(vector<vector<char> > &grid) {
            if (grid.empty() || grid[0].empty()) return 0;
            int m = grid.size(), n = grid[0].size(), res = 0;
            vector<vector<bool> > visited(m, vector<bool>(n, false));
            for (int i = 0; i < m; ++i) {
                for (int j = 0; j < n; ++j) {
                    if (grid[i][j] == '1' && !visited[i][j]) {
                        numIslandsDFS(grid, visited, i, j);
                        ++res;
                    }
                }
            }
            return res;
        }
        void numIslandsDFS(vector<vector<char> > &grid, vector<vector<bool> > &visited, int x, int y) {
            if (x < 0 || x >= grid.size()) return;
            if (y < 0 || y >= grid[0].size()) return;
            if (grid[x][y] != '1' || visited[x][y]) return;
            visited[x][y] = true;
            numIslandsDFS(grid, visited, x - 1, y);
            numIslandsDFS(grid, visited, x + 1, y);
            numIslandsDFS(grid, visited, x, y - 1);
            numIslandsDFS(grid, visited, x, y + 1);
        }
    };
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  • 原文地址:https://www.cnblogs.com/strawqqhat/p/10602370.html
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