zoukankan      html  css  js  c++  java
  • [LeetCode#130]Surrounded Regions

    Problem:

    Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'.

    A region is captured by flipping all 'O's into 'X's in that surrounded region.

    For example,

    X X X X
    X O O X
    X X O X
    X O X X
    

    After running your function, the board should be:

    X X X X
    X X X X
    X X X X
    X O X X

    Analysis:

    Since we have already solved the problem of isolated islands, this problem is very easy to think about.
    Instant idea:
    We begain from each 'O' to merge, once we encounter a 'O' at the broder, all elements this DFS process would not be captured. It seems reasonable, but we have four directions to streach, to pass the state of whether a element meet the border could be a nightmare for implementation.
    
    Think the problem through other way:
    Since we decide wether a region could be captured based on whether there is a island on the border. Whey not we start from each island at border, then try to reach other islands connected with it. After the process, the islands that were not connected would be captured. 
    Step1 : start from border elements.
    for (int i = 0; i < row_len; i++) {
        if (board[i][0] == 'O')
            merge(board, i, 0);
        if (board[i][column_len-1] == 'O')
            merge(board, i, column_len-1);
                
    }
    
    Step2 : use DFS to reach all elementes in the same region, and tag them with '#' tag.
    private void merge(char[][] board, int i, int j) {
        if (i < 0 || j < 0 || i > board.length - 1 || j > board[0].length -1)
            return;
        if (board[i][j] == 'X' || board[i][j] == '#') return;
            board[i][j] = '#';
        merge(board, i-1, j);
        merge(board, i+1, j);
        merge(board, i, j-1);
        merge(board, i, j+1);
    }
    
    Step3 : check all elements on the board, if a island was not tagged, it should be caputured.
    private void merge(char[][] board, int i, int j) {
        if (i < 0 || j < 0 || i > board.length - 1 || j > board[0].length -1)
            return;
        if (board[i][j] == 'X' || board[i][j] == '#') return;
            board[i][j] = '#';
        merge(board, i-1, j);
        merge(board, i+1, j);
        merge(board, i, j-1);
        merge(board, i, j+1);
    }
    
    
    Solution 1: (DFS)
    public class Solution {
        public void solve(char[][] board) {
            if (board == null || board.length == 0 || board[0].length == 0)
                return;
            int row_len = board.length;
            int column_len = board[0].length;
            for (int i = 0; i < row_len; i++) {
                if (board[i][0] == 'O')
                    merge(board, i, 0);
                if (board[i][column_len-1] == 'O')
                    merge(board, i, column_len-1);
                
            }
            for (int j = 0; j < column_len; j++) {
                if (board[0][j] == 'O')
                    merge(board, 0, j);
                if (board[row_len-1][j] == 'O')
                    merge(board, row_len-1, j);
            }
            for (int i = 0; i < row_len; i++) {
                for (int j = 0; j < column_len; j++) {
                    if (board[i][j] == '#')
                        board[i][j] = 'O';
                    else 
                        board[i][j] = 'X';
                }
            }
        }
    
        private void merge(char[][] board, int i, int j) {
            if (i < 0 || j < 0 || i > board.length - 1 || j > board[0].length -1)
                return;
            if (board[i][j] == 'X' || board[i][j] == '#') return;
            board[i][j] = '#';
            merge(board, i-1, j);
            merge(board, i+1, j);
            merge(board, i, j-1);
            merge(board, i, j+1);
        }
    }
    
    Even though the above solution is right, it could easily incure stack overflow prolems.
    We should try to replace the DFS method to BFS method to avoid recursive calls. 
    
    Skill:
    Rather than tag a island, and then use dfs search along one direction.
    
    private void merge(char[][] board, int i, int j) {
            if (i < 0 || j < 0 || i > board.length - 1 || j > board[0].length -1)
                return;
            if (board[i][j] == 'X' || board[i][j] == '#') return;
            board[i][j] = '#';
            merge(board, i-1, j);
            ...
    }
    
    We could separte the tag process and search process to avoid recursive. 
    The bridge between those twp steps are a queue.
    step1: tag a island  => put the island into the queue
    private void fill(char[][] board, int i, int j, Queue<Integer> queue) {
        int m = board.length;
        int n = board[0].length;
        if (i < 0 || j < 0 || i >= m || j >= n || board[i][j] != 'O')
            return;
        queue.offer(i*n + j);
        board[i][j] = '#';
    }
    
    step2: dequeue a island from the queue => tag four islands around it. 
    private void bfs(char[][] board, int i, int j, Queue<Integer> queue) {
        int n = board[0].length;
        //the start element of BFS search
        fill(board, i, j, queue);
        while (!queue.isEmpty()) {
            int index = queue.poll();
            int x = index / n;
            int y = index % n;
            fill(board, x, y-1, queue);
            fill(board, x, y+1, queue);
            fill(board, x-1, y, queue);
            fill(board, x+1, y, queue);
        }
    }
    
    Thus we could avoid the recursive of tagging a island and serach a direction of the island. all directions' island was put into the queue at the same call. 
    
    Note: We should still use the checking condition we used in DFS, only island == '0' could be tagged. and the direction around it should be searched. 
    if (i < 0 || j < 0 || i >= m || j >= n || board[i][j] != 'O')
        return;
    Key: board[i][j] != 'O' at here also means 
    iff boad[i][j] == '#', the island has already been tagged, it should not visited again. This case is very common in BFS searching.

    Solution:

    public class Solution {
        public void solve(char[][] board) {
            if (board == null || board.length == 0 || board[0].length == 0)
                return;
            Queue<Integer> queue = new LinkedList<Integer> ();
            int row_len = board.length;
            int column_len = board[0].length;
            
            for (int i = 0; i < row_len; i++) {
                if (board[i][0] == 'O')
                    bfs(board, i, 0, queue);
                if (board[i][column_len-1] == 'O')
                    bfs(board, i, column_len-1, queue);
                
            }
            for (int j = 0; j < column_len; j++) {
                if (board[0][j] == 'O')
                    bfs(board, 0, j, queue);
                if (board[row_len-1][j] == 'O')
                    bfs(board, row_len-1, j, queue);
            }
            for (int i = 0; i < row_len; i++) {
                for (int j = 0; j < column_len; j++) {
                    if (board[i][j] == '#')
                        board[i][j] = 'O';
                    else 
                        board[i][j] = 'X';
                }
            }
        }
        
        //separate the process of forking!!!
        private void bfs(char[][] board, int i, int j, Queue<Integer> queue) {
            int n = board[0].length;
            //the start element of BFS search
            fill(board, i, j, queue);
            while (!queue.isEmpty()) {
                int index = queue.poll();
                int x = index / n;
                int y = index % n;
                fill(board, x, y-1, queue);
                fill(board, x, y+1, queue);
                fill(board, x-1, y, queue);
                fill(board, x+1, y, queue);
            }
        }
        
        private void fill(char[][] board, int i, int j, Queue<Integer> queue) {
            int m = board.length;
            int n = board[0].length;
            if (i < 0 || j < 0 || i >= m || j >= n || board[i][j] != 'O')
                return;
            queue.offer(i*n + j);
            board[i][j] = '#';
        }
    }
  • 相关阅读:
    仰视源代码,实现strcmp
    Wicket实战(一)概述
    在弱网传输的情况下,是怎么做到节约流量的(面试小问题,Android篇)
    nRF52832之硬件I2C
    Android开源项目SlidingMenu的学习笔记(一)
    MVC设计模式
    报表应用结构优化之数据分库存储
    extern &quot;C&quot; 的含义:实现C++与C及其他语言的混合编程
    phpstorm改变文件编码由utf变为gbk
    mysql database 格式的查看和改变
  • 原文地址:https://www.cnblogs.com/airwindow/p/4762204.html
Copyright © 2011-2022 走看看