Careercup - Google面试题 - 5727310284062720

2014-05-06 14:04

题目链接

原题：

given an 2D matrix M, is filled either using X or O, you need to find the region which is filled by O and surrounded by X
and fill it with X. 
 
example : 
 
X X X X X
X X O O X
X X O O X
O X X X X 
 
Answer : 
 
X X X X X
X X X X X
X X X X X
O X X X X 
 
example : 
 
X X X X X
X X O O X
X X O O O
O X X X X 
 
answer :
X X X X X
X X O O X
X X O O O
O X X X X

题目：参见Leetcode题目Surrounded Regions。

解法：题解位于LeetCode - Surrounded Regions。

代码：

 // http://www.careercup.com/question?id=5727310284062720
 #include <iostream>
 #include <queue>
 #include <string>
 #include <vector>
 using namespace std;
 
 class Solution {
 public:
     void solve(vector<vector<char> > &board) {
         // Should n or m be smaller than 3, there'll be no captured region.
         n = (int)board.size();
         if (n < ) {
             return;
         }
 
         m = (int)board[].size();
         if (m < ) {
             return;
         }
 
         int i, j;
 
         // if an 'O' is on the border, all of its connected 'O's are not captured.
         // so we scan the border and mark those 'O's as free.
 
         // the top row
         for (j = ; j < m; ++j) {
             if (board[][j] == 'O') {
                 check_region(board, , j);
             }
         }
 
         // the bottom row
         for (j = ; j < m; ++j) {
             if (board[n - ][j] == 'O') {
                 check_region(board, n - , j);
             }
         }
 
         // the left column
         for (i = ; i < n - ; ++i) {
             if (board[i][] == 'O') {
                 check_region(board, i, );
             }
         }
 
         // the right column
         for (i = ; i < n - ; ++i) {
             if (board[i][m - ] == 'O') {
                 check_region(board, i, m - );
             }
         }
 
         // other unchecked 'O's are all captured
         for (i = ; i < n; ++i) {
             for (j = ; j < m; ++j) {
                 if (board[i][j] == '#') {
                     // free 'O's
                     board[i][j] = 'O';
                 } else if (board[i][j] == 'O') {
                     // captured 'O's
                     board[i][j] = 'X';
                 }
             }
         }
     }
 private:
     int n, m;
 
     void check_region(vector<vector<char> > &board, int startx, int starty) {
         if (startx <  || startx > n -  || starty <  || starty > m - ) {
             return;
         }
         if (board[startx][starty] == 'O') {
             board[startx][starty] = '#';
             check_region(board, startx - , starty);
             check_region(board, startx + , starty);
             check_region(board, startx, starty - );
             check_region(board, startx, starty + );
         }
     }
 };
 
 int main()
 {
     int n, m;
     int i, j;
     string str;
     vector<vector<char> > board;
     Solution sol;
 
     while (cin >> n >> m && (n >  && m > )) {
         board.resize(n);
         for (i = ; i < n; ++i) {
             cin >> str;
             board[i].resize(m);
             for (j = ; j < m; ++j) {
                 board[i][j] = str[j];
             }
         }
         sol.solve(board);
         for (i = ; i < n; ++i) {
             for (j = ; j < m; ++j) {
                 cout << board[i][j];
             }
             cout << endl;
         }
 
         for (i = ; i < n; ++i) {
             board[i].clear();
         }
         board.clear();
     }
 
     return ;
 }