标签:memory man modifying col other empty boa leshi follow
Given an 2D board, count how many battleships are in it. The battleships are represented with ‘X‘
s, empty slots are represented with‘.‘
s. You may assume the following rules:
1xN
(1 row, N columns) or Nx1
(N rows, 1 column), where N can be of any size.Example:
X..X ...X ...X
In the above board there are 2 battleships.
Invalid Example:
...X XXXX ...X
This is an invalid board that you will not receive - as battleships will always have a cell separating between them.
Follow up:
Could you do it in one-pass, using only O(1) extra memory and without modifying the value of the board?
class Solution { public: int countBattleships(vector<vector<char>>& board) { int count = 0; for(int i=0;i<board.size();i++) for(int j=0;j<board[0].size();j++) if(board[i][j]==‘X‘ && (i==0 || board[i-1][j]!=‘X‘) && (j==0 || board[i][j-1]!=‘X‘)) count++; return count; } };
标签:memory man modifying col other empty boa leshi follow
原文地址:http://www.cnblogs.com/wujufengyun/p/7462424.html