leetcode[C++]Surrounded Regions
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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
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
题目大意:给定一个矩形,将‘X’字符包围的‘O’字符变成‘X’字符。
解题思路:首先,当矩形的大小小于等于2x2时,矩形是无需改变的,直接返回。其次,我们应该从矩形的边缘开始搜索,即在矩形边缘(矩形最外层的四条边)上的字符‘O’以及和边缘上的字符挨着(所谓挨着,即从上下左右四个方向的任意一个方向出发可以到达该字符,且路径上全是‘O’字符)的字符‘O’一定不会被包围。综上,我们可以从矩形的边缘出发,标记出不能被包围的‘O’字符,最后遍历一遍矩形把被包围的‘O’字符变成‘X’,而被标记的字符还原成‘O’字符。我们可以使用DFS和BFS来标记没有被包围的‘O’字符。
AC代码:
class Solution {public: void helper(vector<vector<char>>&board,int i,int j)//dfs搜索 { if(i>1) { if(board[i][j]=='O') { board[i][j]='*'; helper(board,i-1,j); board[i][j]='O';//这里为什么要在还原成‘O’字符,是因为如果从i,j出发可能四个方向都可以扩展,如果此处不还原成‘O’字符,会影响到其他方向的扩展。 } } if(i<board.size()-1) { if(board[i][j]=='O') { board[i][j]='*'; helper(board,i+1,j); board[i][j]='O'; } } if(j>1) { if(board[i][j]=='O') { board[i][j]='*'; helper(board,i,j-1); board[i][j]='O'; } } if(j<board[0].size()-1) { if(board[i][j]=='O') { board[i][j]='*'; helper(board,i,j+1); board[i][j]='O'; } } if(board[i][j]=='O') { board[i][j]='*'; }//四个方向都扩展完以后,重新标记该字符。 } void solve(vector<vector<char>>& board) { int row=board.size(); if(row==0) return; int col=board[0].size(); if(row <=2 || col <=2) return; //直接返回。 for(int i=0;i!=row;++i) { helper(board,i,0); helper(board,i,col-1); } for(int j=0;j!=col;++j) { helper(board,0,j); helper(board,row-1,j); } //对边缘开始搜索。 for(int i=0;i!=row;++i) for(int j=0;j!=col;++j) { if(board[i][j]=='*') board[i][j]='O'; else if(board[i][j]=='O') { board[i][j]='X'; } } }};
本解法的dfs代码显得比较冗余,应该可以做一些简化,网上有类似如下的dfs搜索代码,但我在leetcode跑的时候,会报runtime error,个人觉得可能是递归层次太多造成的,代码如下:
void searchMark(int x, int y, vector<vector<char>> &board){ if(x < 0 || x >= board.size() || y < 0 || y >= board[x].size()) return; if(board[x][y] == 'O'){ board[x][y] = '1'; searchMark(x, y - 1, board); searchMark(x, y + 1, board); searchMark(x + 1, y, board); searchMark(x - 1, y, board); } }
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