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leetcode笔记（七）529. Minesweeper

2018-07-03 01:01:17来源：博客园阅读 ()

题目描述

Let's play the minesweeper game (Wikipedia, online game)!

You are given a 2D char matrix representing the game board. 'M' represents an unrevealed mine, 'E' represents an unrevealed empty square, 'B' represents a revealed blank square that has no adjacent (above, below, left, right, and all 4 diagonals) mines, digit ('1' to '8') represents how many mines are adjacent to this revealed square, and finally 'X' represents a revealed mine.

Now given the next click position (row and column indices) among all the unrevealed squares ('M' or 'E'), return the board after revealing this position according to the following rules:

If a mine ('M') is revealed, then the game is over - change it to 'X'.
If an empty square ('E') with no adjacent mines is revealed, then change it to revealed blank ('B') and all of its adjacent unrevealed squares should be revealed recursively.
If an empty square ('E') with at least one adjacent mine is revealed, then change it to a digit ('1' to '8') representing the number of adjacent mines.
Return the board when no more squares will be revealed.

Example 1:

Input: 

[['E', 'E', 'E', 'E', 'E'],
 ['E', 'E', 'M', 'E', 'E'],
 ['E', 'E', 'E', 'E', 'E'],
 ['E', 'E', 'E', 'E', 'E']]

Click : [3,0]

Output: 

[['B', '1', 'E', '1', 'B'],
 ['B', '1', 'M', '1', 'B'],
 ['B', '1', '1', '1', 'B'],
 ['B', 'B', 'B', 'B', 'B']]

Explanation:

Example 2:

Input: 

[['B', '1', 'E', '1', 'B'],
 ['B', '1', 'M', '1', 'B'],
 ['B', '1', '1', '1', 'B'],
 ['B', 'B', 'B', 'B', 'B']]

Click : [1,2]

Output: 

[['B', '1', 'E', '1', 'B'],
 ['B', '1', 'X', '1', 'B'],
 ['B', '1', '1', '1', 'B'],
 ['B', 'B', 'B', 'B', 'B']]

Explanation:

Note:

The range of the input matrix's height and width is [1,50].
The click position will only be an unrevealed square ('M' or 'E'), which also means the input board contains at least one clickable square.
The input board won't be a stage when game is over (some mines have been revealed).
For simplicity, not mentioned rules should be ignored in this problem. For example, you don't need to reveal all the unrevealed mines when the game is over, consider any cases that you will win the game or flag any squares.

解题思路：

题目属于背景复杂，但本质不难的类型。因为没玩过扫雷游戏，所以看了半天题目，了解了规则后，可以发现属于传统的广度优先搜索问题。基础实现的时候，会超时间。

最后改了半天边缘问题，仍然是超时。最后，看了英文版leetcode的讨论区，发现就一行神秘的代码，就可以解决重复处理节点的问题～～～

示例代码

class Solution {
public:
    vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) {
        int row = click[0];
        int col = click[1];
        
        int maxRow = board.size();
        int maxCol = board[0].size();
        vector<vector<char>> results;
        for(int i = 0; i < maxRow; i++)
        {
            vector<char> temp(board[i]);
            results.push_back(temp);
        }
        // M
        if(results[row][col] == 'M')
            results[row][col] = 'X';
        // E
        else if(results[row][col] == 'E')
        {
            queue<pair<int, int>> eNodes;
            eNodes.push(make_pair(row, col));
            while(!eNodes.empty())
            {
                pair<int, int> curr = eNodes.front();
                eNodes.pop();
                row = curr.first;
                col = curr.second;
                results[row][col] = '0';
                // check neighbors
                for(int i = -1; i <= 1; i++)
                {
                    int currRow = row + i;
                    if(currRow < maxRow && currRow >= 0)
                    {
                        for(int j = -1; j <= 1; j++)
                        {
                            if(i == 0 && j == 0)
                                continue;
                            int currCol = col + j;
                            if(currCol < maxCol && currCol >= 0)
                            {
                                // M add to counter
                                if(results[currRow][currCol] == 'M')
                                {
                                    results[row][col]++;
                                }
                            }
                        }
                    }
                }
                // around no M
                if(results[row][col] == '0')
                {
                    results[row][col] = 'B';
                    // add neighbors to deal
                    for(int i = -1; i <= 1; i++)
                    {
                        int currRow = row + i;
                        if(currRow < maxRow && currRow >= 0)
                        {
                            for(int j = -1; j <= 1; j++)
                            {
                                if(i == 0 && j == 0)
                                    continue;
                                int currCol = col + j;
                                if(currCol < maxCol && currCol >= 0)
                                {
                                    // E add to deal
                                    if(results[currRow][currCol] == 'E')
                                    {
                                        eNodes.push(make_pair(currRow, currCol));
                                        results[currRow][currCol] = 'B'; // optimize to avoid repeatly added
                                    }
                                }
                            }
                        }
                    }
                }
            }// end of dealing
        }
        
        
        return results;
            
    }
};