Given an 2D board, count how many different battleships are in it. The battleships are represented with '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) orNx1(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.
找到有多少条船,横向或纵向不相隔的X组成一条船,船之间至少有一个空点
解题思路:
遍历二维数组,每遍历到一个位置的时候,只需要考虑其左边和上面的位置是否为X即可
1 int countBattleships(char** board, int boardRowSize, int boardColSize) { 2 int i,j; 3 int count=0; 4 for(i=0;i<boardRowSize;i++) 5 for(j=0;j<boardColSize;j++) 6 if(board[i][j]=='X') 7 { 8 if(i==0) 9 { 10 if(j==0||board[i][j-1]=='.') 11 count++; 12 } 13 else 14 { 15 if((j==0&&board[i-1][j]=='.')||(board[i-1][j]=='.'&&board[i][j-1]=='.')) 16 count++; 17 } 18 } 19 return count; 20 } 21 //遍历到每个元素时候,只需要考虑其前面和上面是否为'.'即可