标签:tween 组成 aced pre val rate repr empty shape
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:
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.
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
标签:tween 组成 aced pre val rate repr empty shape
原文地址:http://www.cnblogs.com/fcyworld/p/6146593.html