leetcode Surrounded Regions
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题意:给定一个2维字符数组,由字符'X'和'O'组成,找出所有被包围的'O'(O向左,向右,向上,向下走在走出边界前会遇到字符'X'),将这些被包围的'O'变换成为'X'。
思路:我们可以将2维数组的4个边界看成4个墙壁,从墙壁每一个字符为'O'处出发搜索,将每一个遇到的'O'做标记,直到越界或者遇见'X'的时候停止。然后重新遍历,将做标记的'O'处变换。
代码实现:
class Solution {public: void search(vector<vector<char>> &board, int i, int j) { if(i < 0 || i >= board[0].size() || j < 0 || j >= board.size()) { return ; } if(board[i][j] == 'O') { board[i][j] = '#'; } else { return; } search(board, i + 1, j); search(board, i - 1, j); search(board, i, j + 1); search(board, i, j - 1); } void solve(vector<vector<char>> &board) { // Start typing your C/C++ solution below // DO NOT write int main() function if(board.size() == 0) { return ; } int len1 = board[0].size(); int len2 = board.size(); for(int i = 0; i < len1; i++) { search(board, 0, i); search(board, len2 - 1, i); } for(int i = 0; i < len2; i ++) { search(board, i, 0); search(board, i, len1 - 1); } for(int i = 0; i < len1; ++i) { for(int j = 0; j < len2; ++j) { if(board[i][j] == '#') { board[i][j] = 'O'; } else if(board[i][j] == 'O') { board[i][j] = 'X'; } } } return; }};
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