leetcode 130 —— Sum Root to Leaf Numbers

最新推荐文章于 2023-10-10 09:42:58 发布

自犬邦

最新推荐文章于 2023-10-10 09:42:58 发布

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分类专栏：算法学习数据结构

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本文介绍了一种解决二维棋盘中被X包围的O区域捕获问题的算法。通过遍历棋盘，使用暴力法检查每个'O'是否被完全包围，并将符合条件的'O'标记为'X'，最终实现对被包围区域的有效捕获。

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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

思路1：暴力法，343ms，差一点超时

class Solution {
public:
	void solve(vector<vector<char>>& board) {
		bool flag = true;
		int m = board.size();
		if (m == 0) return;
		int n = board[0].size();
		for (int i = 0; i < m; i++){
			for (int j = 0; j < n; j++){
				if (board[i][j] == 'X')
					continue;
				if (board[i][j] == 'O'){
					if (check(board, i, j)){
						board[i][j] = 'Y';
					}
					//	else{
					//		recover(board, i, j);
					//	}
				}
			}
		}
		while (flag){
			flag = false;
			for (int i = 0; i < m; i++){
				for (int j = 0; j < n; j++){
					if (board[i][j] == 'O'){
						recover(board, i, j,flag);
					}
				}
			}
		}
		for (int i = 0; i < m; i++){
			for (int j = 0; j < n; j++){
				if (board[i][j] == 'Y')
					board[i][j] = 'X';
			}
		}
	}
	bool check(vector<vector<char>>& board, int i, int j){
		bool left, right, up, down;
		left = right = up = down = false;

		for (int m = i - 1; m >= 0; m--){
			if (board[m][j] == 'X'){
				left = true;
				break;
			}
		}
		for (int m = i + 1; m < board.size(); m++){
			if (board[m][j] == 'X'){
				right = true;
				break;
			}
		}
		for (int m = j - 1; m >= 0; m--){
			if (board[i][m] == 'X'){
				up = true;
				break;
			}
		}
		for (int m = j + 1; m < board[0].size(); m++){
			if (board[i][m] == 'X'){
				down = true;
				break;
			}
		}
		return left&&right&&up&&down;
	}
	//把Y变成O
	void recover(vector<vector<char>>& board, int i, int j,bool& flag){  
		for (int m = i - 1; m >= 0; m--){
			if (board[m][j] == 'Y'){
				board[m][j] = 'O';
				flag = true;
			}
			if (board[m][j] == 'X')
				break;
		}
		for (int m = i + 1; m < board.size(); m++){
			if (board[m][j] == 'Y'){
				board[m][j] = 'O';
				flag = true;
			}
			if (board[m][j] == 'X')
				break;
		}
		for (int m = j - 1; m >= 0; m--){
			if (board[i][m] == 'Y'){
				board[i][m] = 'O';
				flag = true;
			}
			if (board[i][m] == 'X')
				break;
		}
		for (int m = j + 1; m < board[0].size(); m++){
			if (board[i][m] == 'Y'){
				flag = true;
				board[i][m] = 'O';
			}
			if (board[i][m] == 'X')
				break;
		}
	}  //把Y变成O
};