zoukankan      html  css  js  c++  java
  • LeetCode

    Surrounded Regions

    2014.2.27 00:59

    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

    Solution:

      This problem is a test on Flood Fill Algorithm. You can do it with DFS or BFS, but the latter will result in TLE.

      Why is BFS so slow? Because the queue operation is not strictly O(1), but rather ammortized to O(1). The bad case for some of the queue operations will be O(n). While using BFS you don't have to worry about call stack overflow, but rather the speed of a queue.

      Total time complexity is O(n^2). Space complexity is O(n^2), mainly from parameters in recursive function calls.

    Accepted code:

     1 // 1CE, 3TLE, 1AC, DFS accepted but BFS failed?
     2 #include <queue>
     3 #include <vector>
     4 using namespace std;
     5 
     6 class Solution {
     7 public:
     8     void solve(vector<vector<char> > &board) {
     9         // Should n or m be smaller than 3, there'll be no captured region.
    10         n = (int)board.size();
    11         if (n < 3) {
    12             return;
    13         }
    14         
    15         m = (int)board[0].size();
    16         if (m < 3) {
    17             return;
    18         }
    19         
    20         int i, j;
    21         
    22         // if an 'O' is on the border, all of its connected 'O's are not captured.
    23         // so we scan the border and mark those 'O's as free.
    24         
    25         // the top row
    26         for (j = 0; j < m; ++j) {
    27             if (board[0][j] == 'O') {
    28                 check_region(board, 0, j);
    29             }
    30         }
    31         
    32         // the bottom row
    33         for (j = 0; j < m; ++j) {
    34             if (board[n - 1][j] == 'O') {
    35                 check_region(board, n - 1, j);
    36             }
    37         }
    38         
    39         // the left column
    40         for (i = 1; i < n - 1; ++i) {
    41             if (board[i][0] == 'O') {
    42                 check_region(board, i, 0);
    43             }
    44         }
    45         
    46         // the right column
    47         for (i = 1; i < n - 1; ++i) {
    48             if (board[i][m - 1] == 'O') {
    49                 check_region(board, i, m - 1);
    50             }
    51         }
    52         
    53         // other unchecked 'O's are all captured
    54         for (i = 0; i < n; ++i) {
    55             for (j = 0; j < m; ++j) {
    56                 if (board[i][j] == '#') {
    57                     // free 'O's
    58                     board[i][j] = 'O';
    59                 } else if (board[i][j] == 'O') {
    60                     // captured 'O's
    61                     board[i][j] = 'X';
    62                 }
    63             }
    64         }
    65     }
    66 private:
    67     int n, m;
    68     
    69     void check_region(vector<vector<char> > &board, int startx, int starty) {
    70         if (startx < 0 || startx > n - 1 || starty < 0 || starty > m - 1) {
    71             return;
    72         }
    73         if (board[startx][starty] == 'O') {
    74             board[startx][starty] = '#';
    75             check_region(board, startx - 1, starty);
    76             check_region(board, startx + 1, starty);
    77             check_region(board, startx, starty - 1);
    78             check_region(board, startx, starty + 1);
    79         }
    80     }
    81 };
  • 相关阅读:
    所有者权益
    金融工具
    或有事项
    股份支付
    借款费用
    Keycode对照表
    js(jQuery)获取时间搜集
    jQuery实现CheckBox全选、全不选
    JS 截取字符串函数
    jQuery mouseenter与mouseleave
  • 原文地址:https://www.cnblogs.com/zhuli19901106/p/3570554.html
Copyright © 2011-2022 走看看