dfs例题 130. 被围绕的区域

dfs例题

:

130. 被围绕的区域

给你一个 m x n 的矩阵 board ，由若干字符 X 和 O ，找到所有被 X 围绕的区域，并将这些区域里所有的 O 用 X 填充。

示例 1：

输入：board = [["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]
输出：[["X","X","X","X"],["X","X","X","X"],["X","X","X","X"],["X","O","X","X"]]
解释：被围绕的区间不会存在于边界上，换句话说，任何边界上的 O 都不会被填充为 X。

任何不在边界上，或不与边界上的 O 相连的 O 最终都会被填充为 X。如果两个元素在水平或垂直方向相邻，则称它们是“相连”的。

示例 2：

输入：board = [["X"]]
输出：[["X"]]

编辑于just now

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code.cpp919字节

class Solution {
public:
    int m , n;
    void dfs(int x, int y,vector<vector<char>>& board){
        if(x<0||y<0||x>=m||y>=n||board[x][y]!=O){
            return;
        }
        board[x][y] = A;
        dfs(x-1, y, board);
        dfs(x+1, y, board);
        dfs(x, y-1, board);
        dfs(x, y+1, board);
    }
    void solve(vector<vector<char>>& board) {
        m = board.size(); n = board[0].size();
        for(int i = 0; i < n; i++){
            dfs(0, i, board);
            dfs(m-1, i, board);
        }
        for(int i = 0; i < m; i++){
            dfs(i, 0, board);
            dfs(i, n-1, board);
        }
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(board[i][j]==A){
                    board[i][j] = O;
                }else if(board[i][j]==O){
                    board[i][j] = X; 
                }
            }
        }
    }
};