lintcode - 被围绕的区域

 1 class Solution {
 2 public:
 3     /*
 4      * @param board: board a 2D board containing 'X' and 'O'
 5      * @return: nothing
 6      */
 7     bool flag[220][220];
 8     vector<pair<int, int>> v;
 9     const int dir[4][2] = {{0,1}, {1,0}, {0, -1}, {-1, 0}};
10     int isChange = 1;
11     void change_board(vector<vector<char>> &board){
12         for(size_t i = 0; i < v.size(); ++i){
13                 int x = v[i].first;
14                 int y = v[i].second;
15                 board[x][y] = 'X';
16         }
17     }
18     void surroundedRegions(vector<vector<char>> &board) {
19         // write your code here
20         memset(flag, 0, sizeof(flag));
21         for(size_t i = 0; i < board.size(); ++i){
22                 for(size_t j = 0; j < board[i].size(); ++j){
23                         if(flag[i][j] == 0 && board[i][j] == 'O'){
24                                 v.push_back(make_pair(i,j));
25                                 bfs(i,j, board);
26                                 if(isChange == 1){
27                                         //printf("change:%d %d\n",i,j);
28                                         change_board(board);
29                                 }
30                                 v.clear();
31                                 isChange = 1;
32                         }
33                 }
34         }
35     }
36     void bfs(int x, int y, vector<vector<char>> board){
37         queue<pair<int, int>> que;
38         que.push(make_pair(x,y));
39         flag[x][y] = 1;
40         while(!que.empty()){
41                 pair<int, int> p = que.front();
42                 que.pop();
43 
44 
45 
46                 int nx = 0, ny = 0;
47                 for(int i = 0; i < 4; ++i){
48                         nx = p.first + dir[i][0];
49                         ny = p.second + dir[i][1];
50                         if(nx >= 0 && nx < board.size() && ny >= 0 && ny < board[0].size()){
51                                 if(flag[nx][ny] == 0 && board[nx][ny] == 'O'){
52                                         v.push_back(make_pair(nx, ny));
53                                         que.push(make_pair(nx, ny));
54                                         flag[nx][ny] = 1;
55                                 }
56                         } else {
57                                // printf("dsad:%d %d\n",nx, ny);
58                                 isChange = 0;
59                         }
60 
61                 }
62         }
63     }
64 };

dfs会爆栈 可以 bfs或者用栈模拟函数

posted @ 2017-09-18 20:09 GeniusYang 阅读(...) 评论(...) 编辑 收藏