zoukankan      html  css  js  c++  java
  • [LC] 200. Number of Islands

    Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

    Example 1:

    Input:
    11110
    11010
    11000
    00000
    
    Output: 1
    

    Example 2:

    Input:
    11000
    11000
    00100
    00011
    
    Output: 3

    Solution 1:
    Time: O(M * N)
    class Solution {
        public int numIslands(char[][] grid) {
            if (grid == null || grid.length == 0 || grid[0].length == 0) {
                return 0;
            }
            int row = grid.length;
            int col = grid[0].length; 
            int res = 0;
            for (int i = 0; i < row; i++) {
                for (int j = 0; j < col; j++) {
                    if (grid[i][j] == '1') {
                        floodFill(grid, i, j);
                        res += 1;
                    }
                }
            }
            return res;
        }
        
        private void floodFill(char[][] grid, int i, int j) {
            // need to include grid[i][j] == 0 in base case
            if (i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == '0') {
                return;
            }
            grid[i][j] = '0';
            floodFill(grid, i - 1, j);
            floodFill(grid, i + 1, j);
            floodFill(grid, i, j - 1);
            floodFill(grid, i, j + 1);  
        }
    }

    solution 2:

    class Solution {
        private int row;
        private int col;
        public int numIslands(char[][] grid) {
            if (grid == null || grid.length == 0 || grid[0].length == 0) {
                return 0;
            }
            row = grid.length;
            col = grid[0].length; 
            int res = 0;
            for (int i = 0; i < row; i++) {
                for (int j = 0; j < col; j++) {
                    if (grid[i][j] == '1') {
                        floodFillBFS(grid, i, j);
                        res += 1;
                    }
                }
            }
            return res;
        }
        
        private void floodFillBFS(char[][] grid, int i, int j) {
            grid[i][j] = '0';
            Queue<Integer> queue = new LinkedList<>();
            queue.offer(i * col + j);
            while (!queue.isEmpty()) {
                int cur = queue.poll();
                int m = cur / col;
                int n = cur % col;
                if (m - 1 >= 0 && grid[m - 1][n] == '1') {
                    queue.offer((m - 1) * col + n);
                    grid[m - 1][n] = '0';
                }
                if (m + 1 < row && grid[m + 1][n] == '1') {
                    queue.offer((m + 1) * col + n);            
                    grid[m + 1][n] = '0';
                }
                if (n - 1 >= 0 && grid[m][n - 1] == '1') {
                    queue.offer(m * col + n - 1);
                    grid[m][n - 1] = '0';
                }
                if (n + 1 < col && grid[m][n + 1] == '1') {
                    queue.offer(m * col + n + 1);
                    grid[m][n + 1] = '0';
                }    
            }
        }
    }

    Solution 3:

    class Solution {
        private int row;
        private int col;
        public int numIslands(char[][] grid) {
            if (grid == null || grid.length == 0 || grid[0].length == 0) {
                return 0;
            }
            row = grid.length;
            col = grid[0].length; 
            int res = 0;
            for (int i = 0; i < row; i++) {
                for (int j = 0; j < col; j++) {
                    if (grid[i][j] == '1') {
                        floodFillBFSOpt(grid, i, j);
                        res += 1;
                    }
                }
            }
            return res;
        }
        
        private void floodFillBFSOpt(char[][] grid, int i, int j) {
            grid[i][j] = '0';
            int[][] directions = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
            Queue<Cell> queue = new LinkedList<>();
            queue.offer(new Cell(i, j));
            while (!queue.isEmpty()) {
                Cell cur = queue.poll();
                for (int[] direction: directions) {
                    int x = cur.getX() + direction[0];
                    int y = cur.getY() + direction[1];
                    if (x >= 0 && x < row && y >= 0 && y < col && grid[x][y] == '1') {
                        queue.offer(new Cell(x, y));
                        grid[x][y] = '0';
                    }
                }
            }
        }
    }
    
    class Cell {
        private int x;
        private int y;
        public Cell(int x, int y) {
            this.x = x;
            this.y = y;
        }
        public int getX() {
            return x;
        }
        
        public int getY() {
            return y;
        }
    }
  • 相关阅读:
    intel instruction 指令速查
    WinDbg双机调试配置
    MSDN上关于WinDbg的手册
    build temu error about SDL
    taintCheck的实现
    Vim使用taglist功能
    Windows编写driver
    cabal替代脚本
    怎样理解Functor与Monad
    haskell基本语法
  • 原文地址:https://www.cnblogs.com/xuanlu/p/11988532.html
Copyright © 2011-2022 走看看