419. Battleships in a Board
2017-05-29 14:36
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Given an 2D board, count how many battleships are in it. The battleships are represented with
empty slots are represented with
You receive a valid board, made of only battleships or empty slots.
Battleships can only be placed horizontally or vertically. In other words, they can only be made of the shape
row, N columns) or
At least one horizontal or vertical cell separates between two battleships - there are no adjacent battleships.
Example:
In the above board there are 2 battleships.
Invalid Example:
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?
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class Solution {
public:
int countBattleships(vector<vector<char>>& board) {
int count=0,R=board.size(),C=board[0].size();
for(int i = 0;i<R;i++){
for(int j = 0;j<C;j++){
if(board[i][j]=='X'&&(i==0||board[i-1][j]!='X')&&(j==0||board[i][j-1]!='X')) count++;
}
}
return count;
}
};
'X's,
empty slots are represented with
'.'s. You may assume the following rules:
You receive a valid board, made of only battleships or empty slots.
Battleships can only be placed horizontally or vertically. In other words, they can only be made of the shape
1xN(1
row, N columns) or
Nx1(N rows, 1 column), where N can be of any size.
At least one horizontal or vertical cell separates between two battleships - there are no adjacent battleships.
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?
Subscribe to see which companies asked this question.
class Solution {
public:
int countBattleships(vector<vector<char>>& board) {
int count=0,R=board.size(),C=board[0].size();
for(int i = 0;i<R;i++){
for(int j = 0;j<C;j++){
if(board[i][j]=='X'&&(i==0||board[i-1][j]!='X')&&(j==0||board[i][j-1]!='X')) count++;
}
}
return count;
}
};
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