. - 力扣(LeetCode)
题目简述:扫雷,点击一个格子,返回整个地图的下一个状态。
对于边界元素,可以设置两个数组,index_row,index_col,遍历到一个格子需要搜索其周围格子时,使当前格子的值一次加上数组的值,如果发现加上之后的值越界了则跳过此格子。
这样写很清晰,节省了很多代码量。
class Solution {
public:vector<vector<int>>hasboom;int index_row[8]={0,0,1,1,1,-1,-1,-1},index_col[8]={-1,1,-1,1,0,-1,1,0};void count(int row,int col){for(int i=0;i<8;i++){int temprow=row+index_row[i],tempcol=col+index_col[i];if(temprow>=0&&temprow<hasboom.size()&&tempcol>=0&&tempcol<hasboom[0].size())hasboom[temprow][tempcol]++;}// if(row>0)// {// hasboom[row-1][col]++;// if(col>0)hasboom[row-1][col-1]++;// if(col<hasboom[0].size()-1)hasboom[row-1][col+1]++;// }// if(row<hasboom.size()-1)// {// hasboom[row+1][col]++;// if(col>0)hasboom[row+1][col-1]++;// if(col<hasboom[0].size()-1)hasboom[row+1][col+1]++;// }// if(col>0)hasboom[row][col-1]++;// if(col<hasboom[0].size()-1)hasboom[row][col+1]++;}vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) {hasboom.resize(board.size());for(int i=0;i<board.size();i++){hasboom[i].resize(board[0].size(),0);}for(int i=0;i<hasboom.size();i++){for(int j=0;j<hasboom[0].size();j++){if(board[i][j]=='M'){count(i,j);}}}queue<pair<int,int>>q;q.emplace(click[0],click[1]);while(!q.empty()){auto[row,col]=q.front();q.pop();if(board[row][col]=='M'){board[row][col]='X';}else if(board[row][col]=='E'){if(hasboom[row][col]==0){board[row][col]='B';for(int index=0;index<8;index++){int temprow=row+index_row[index];int tempcol=col+index_col[index];if(temprow<0||temprow>=board.size()||tempcol<0||tempcol>=board[0].size())continue;else if(board[temprow][tempcol]=='M'||board[temprow][tempcol]=='E')q.emplace(temprow,tempcol);}// if(row>0)// {// if(board[row-1][col]=='M'||board[row-1][col]=='E')// q.emplace(row-1,col);// if(col>0&&(board[row-1][col-1]=='M'||board[row-1][col-1]=='E'))// q.emplace(row-1,col-1);// if(col<board[0].size()-1&&(board[row-1][col+1]=='M'||board[row-1][col+1]=='E'))// q.emplace(row-1,col+1);// }// if(col>0&&(board[row][col-1]=='M'||board[row][col-1]=='E'))// q.emplace(row,col-1);// if(col<board[0].size()-1&&(board[row][col+1]=='M'||board[row][col+1]=='E'))// q.emplace(row,col+1);// if(row<board.size()-1)// {// if(board[row+1][col]=='M'||board[row+1][col]=='E')// q.emplace(row+1,col);// if(col>0&&(board[row+1][col-1]=='M'||board[row+1][col-1]=='E'))// q.emplace(row+1,col-1);// if(col<board[0].size()-1&&(board[row+1][col+1]=='M'||board[row+1][col+1]=='E'))// q.emplace(row+1,col+1);// }}else{board[row][col]='0'+hasboom[row][col];}}}return board;}
};