wap网站建设公司/西安自动seo

一、题目

Given an m x n matrix board containing ‘X’ and ‘O’, capture all regions that are 4-directionally surrounded by ‘X’.

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

Example 1:

Input: board = [[“X”,“X”,“X”,“X”],[“X”,“O”,“O”,“X”],[“X”,“X”,“O”,“X”],[“X”,“O”,“X”,“X”]]
Output: [[“X”,“X”,“X”,“X”],[“X”,“X”,“X”,“X”],[“X”,“X”,“X”,“X”],[“X”,“O”,“X”,“X”]]
Explanation: Notice that an ‘O’ should not be flipped if:

• It is on the border, or
• It is adjacent to an ‘O’ that should not be flipped.
  The bottom ‘O’ is on the border, so it is not flipped.
  The other three ‘O’ form a surrounded region, so they are flipped.
  Example 2:

Input: board = [[“X”]]
Output: [[“X”]]

Constraints:

m == board.length
n == board[i].length
1 <= m, n <= 200
board[i][j] is ‘X’ or ‘O’.

二、题解

visited数组法——使用额外空间

与上题的方法类似，遍历四条边上值为’O’的值，并找出与其相邻的所有’O’，将其标记为已走过，最后遍历整张图，找到没走过的值为’O’的各自，并将其修改为’X’即可。本方法由于使用了visited数组，因此空间复杂度较高。

class Solution {
public:int dirs[4][2] = {1,0,0,1,-1,0,0,-1};void dfs(vector<vector<char>>& board,vector<vector<bool>>& visted,int x,int y){visted[x][y] = true;for(int i = 0;i < 4;i++){int nextX = x + dirs[i][0];int nextY = y + dirs[i][1];if(nextX < 0 || nextX >= board.size() || nextY < 0 || nextY >= board[0].size()) continue;if(board[nextX][nextY] == 'O' && visted[nextX][nextY] == false){visted[nextX][nextY] = true;dfs(board,visted,nextX,nextY);}}}void solve(vector<vector<char>>& board) {int m = board.size();int n = board[0].size();vector<vector<bool>> visted(m,vector<bool>(n,false));//对左和右进行遍历for(int i = 0;i < m;i++){if(board[i][0] == 'O' && visted[i][0] == false) dfs(board,visted,i,0);if(board[i][n-1] == 'O' && visted[i][n-1] == false) dfs(board,visted,i,n-1);}//对上和下进行遍历for(int j = 1;j < n - 1;j++){if(board[0][j] == 'O' && visted[0][j] == false) dfs(board,visted,0,j);if(board[m-1][j] == 'O' && visted[m-1][j] == false) dfs(board,visted,m-1,j);}//改变被包围飞地的值for(int i = 0;i < m;i++){for(int j = 0;j < n;j++){if(board[i][j] == 'O' && visted[i][j] == false) board[i][j] = 'X';}}}
};

不使用额外空间的方法——直接在原数组中使用数字A进行标记

class Solution {
public:int dirs[4][2] = {1,0,0,1,-1,0,0,-1};void dfs(vector<vector<char>>& board,int x,int y){board[x][y] = 'A';for(int i = 0;i < 4;i++){int nextX = x + dirs[i][0];int nextY = y + dirs[i][1];if(nextX < 0 || nextX >= board.size() || nextY < 0 || nextY >= board[0].size()) continue;if(board[nextX][nextY] == 'O'){board[nextX][nextY] = 'A';dfs(board,nextX,nextY);}}}void solve(vector<vector<char>>& board) {int m = board.size();int n = board[0].size();//对左和右进行遍历for(int i = 0;i < m;i++){if(board[i][0] == 'O') dfs(board,i,0);if(board[i][n-1] == 'O') dfs(board,i,n-1);}//对上和下进行遍历for(int j = 1;j < n - 1;j++){if(board[0][j] == 'O') dfs(board,0,j);if(board[m-1][j] == 'O') dfs(board,m-1,j);}//改变被包围飞地的值for(int i = 0;i < m;i++){for(int j = 0;j < n;j++){if(board[i][j] == 'O') board[i][j] = 'X';else if(board[i][j] == 'A') board[i][j] = 'O';}}}
};