Leetcode 130. Surrounded Regions

最新推荐文章于 2023-12-01 14:27:39 发布

Z-Pilgrim

最新推荐文章于 2023-12-01 14:27:39 发布

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分类专栏：搜索-BFS LeetCode题解

本文链接：https://blog.youkuaiyun.com/u011026968/article/details/79029681

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本文介绍了一种使用广度优先搜索(BFS)解决LeetCode围攻区域问题的方法。通过从矩阵的边界开始搜索，标记所有连接到边界的'O'字符，最后将未被标记的'O'字符替换为'X'字符，实现围攻区域的翻转。

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https://leetcode.com/problems/surrounded-regions/description/

我是BFS写的，这里有个小Trick，就是BFS的起点其实可以不是一个点，思路是只要从所有边界点开始BFS就行。代码有点长，但是看了discuss的代码基本思路都一样无非是DFS/BFS

const int SIZE = 2000 + 10;
int visit[SIZE][SIZE];

class Solution {
public:
        
    void solve(vector<vector<char>>& board) {
        if (board.size() == 0) return;
        int n = board.size(), m = board[0].size();
        for (int i = 0; i < board.size(); i++) {
            for (int j = 0; j < board[0].size(); j++) {
                visit[i][j] = 0;
            }
        }
        
        queue < pair<int , int> > q;
        
        for (int i = 0; i < board.size(); i++) {
            if (board[i][0] == 'O') {
                q.push( make_pair( i, 0 ) );
                visit[i][0] = 2; // safe
            } else {
                visit[i][0] = 1; // visit and tobe X
            }
                
            
            if (m != 1 ) {
                if ( board[i][m-1] == 'O' ) {
                    q.push( make_pair(i, m-1) );
                     visit[i][m-1] = 2;
                } else {
                     visit[i][m-1] = 1;
                }
            }
            
        }
        
        for (int i = 1; i < m - 1; i++) {
            if (board[0][i] == 'O') {
                q.push( make_pair( 0, i ) );
                visit[0][i] = 2;
            } else {
                visit[0][i] = 1;
            }
            
            if (n != 1 ) {
                if ( board[n - 1][i] == 'O' ) {
                    q.push( make_pair( n - 1, i ) );
                    visit[n-1][i] = 2;
                } else {
                    visit[n-1][i] = 1;
                }
            }
        }
        
        while (!q.empty()) {
            pair <int, int> tp = q.front();
            q.pop();
            int x = tp.first, y = tp.second;
            if (x -1 >= 0 && !visit[x-1][y] && board[x-1][y] == 'O') {
                visit[x - 1][y] = 2;
                q.push( make_pair(x-1, y) );
            }
            if (x+1 < n && !visit[x+1][y] && board[x+1][y] == 'O') {
                visit[x + 1][y] = 2;
                q.push( make_pair(x+1, y) );
            }
            
            if (y - 1 >= 0 && !visit[x][y-1] && board[x][y-1] == 'O') {
                visit[x][y-1] = 2;
                q.push( make_pair(x, y-1) );
            }
            
            if (y + 1 < m && !visit[x][y+1] && board[x][y+1] == 'O') {
                visit[x][y+1] = 2;
                q.push( make_pair(x, y+1) );
            }
            
        }
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (visit[i][j] != 2 && board[i][j] == 'O' ) {
                    board[i][j] = 'X';
                }
            }
        }
        
    }
};