zoukankan      html  css  js  c++  java
  • [LintCode] Surrounded Regions 包围区域

    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.
    Example

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

    After capture all regions surrounded by 'X', the board should be:

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

    LeetCode上的原题,请参见我之前的博客Surrounded Regions

    解法一:

    class Solution {
    public:
        /**
         * @param board a 2D board containing 'X' and 'O'
         * @return void
         */
        void surroundedRegions(vector<vector<char>>& board) {
            if (board.empty() || board[0].empty()) return;
            int m = board.size(), n = board[0].size();
            for (int i = 0; i < m; ++i) {
                for (int j = 0; j < n; ++j) {
                    if (i == 0 || i == m - 1 || j == 0 || j == n - 1) {
                        if (board[i][j] == 'O') dfs(board, i , j);
                    }
                }
            }
            for (int i = 0; i < m; ++i) {
                for (int j = 0; j < n; ++j) {
                    if (board[i][j] == 'O') board[i][j] = 'X';
                    if (board[i][j] == '$') board[i][j] = 'O';
                }
            }
        }
        void dfs(vector<vector<char>> &board, int x, int y) {
            int m = board.size(), n = board[0].size();
            vector<vector<int>> dir{{0,-1},{-1,0},{0,1},{1,0}};
            board[x][y] = '$';
            for (int i = 0; i < dir.size(); ++i) {
                int dx = x + dir[i][0], dy = y + dir[i][1];
                if (dx >= 0 && dx < m && dy >= 0 && dy < n && board[dx][dy] == 'O') {
                    dfs(board, dx, dy);
                }
            }
        }
    };

    解法二:

    class Solution {
    public:
        /**
         * @param board a 2D board containing 'X' and 'O'
         * @return void
         */
        void surroundedRegions(vector<vector<char>>& board) {
            if (board.empty() || board[0].empty()) return;
            int m = board.size(), n = board[0].size();
            for (int i = 0; i < m; ++i) {
                for (int j = 0; j < n; ++j) {
                    if (i == 0 || i == m - 1 || j == 0 || j == n - 1) {
                        if (board[i][j] == 'O') dfs(board, i , j);
                    }
                }
            }
            for (int i = 0; i < m; ++i) {
                for (int j = 0; j < n; ++j) {
                    if (board[i][j] == 'O') board[i][j] = 'X';
                    if (board[i][j] == '$') board[i][j] = 'O';
                }
            }
        }
        void dfs(vector<vector<char>> &board, int i, int j) {
            int m = board.size(), n = board[0].size();
            if (i < 0 || i >= m || j < 0 || j >= n || board[i][j] != 'O') return;
            board[i][j] = '$';
            dfs(board, i + 1, j);
            dfs(board, i - 1, j);
            dfs(board, i, j + 1);
            dfs(board, i, j - 1);
        }
    };
  • 相关阅读:
    BFPRT(中位数的中位数算法)
    View的事件体系
    Android的消息机制
    BInder机制总结
    Service总结
    IPC机制总结
    AIDL使用绑定启动远程Service出现Service Intent must be explicit: Intent
    线程协作与并发流程控制&AQS
    TCP相关问题整理
    USACO 2015 US OPEN Gold T3: Trapped in the Haybales
  • 原文地址:https://www.cnblogs.com/grandyang/p/5572999.html
Copyright © 2011-2022 走看看