leetcode 419. Battleships in a Board
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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:
- 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...XIn the above board there are 2 battleships.
Invalid Example:
...XXXXX...XThis 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?
我的方法是从左往右,从上往下遍历,遇到遍历过的或'.'就跳掉,遇到'X'就看看到底是往右还是往下发展的。如果是往右发展的,则j++,一直到结束,把battleship+1;如果是往下发展的,则temp=i,temp++,一直到结束,把battleship+1(注意不能i++,因为后几行的列还等着你遍历呢)
public int countBattleships(char[][] board) {int count=0;int xLength=board.length;//xLength是行数int yLength=board[0].length;//yLength是列数int[][] isGoneThroguh = new int[xLength][yLength];for(int i=0;i<xLength;i++){for(int j=0;j<yLength;j++){if(isGoneThroguh[i][j]==1){continue;}if(board[i][j]=='.'){isGoneThroguh[i][j]=1;continue;}if(board[i][j]=='X'){if(j+1<yLength&&board[i][j+1]=='X'){while(j<yLength&&board[i][j]=='X'){isGoneThroguh[i][j]=1;j++;}j--;count++;}else if(i+1<xLength&&board[i+1][j]=='X'){int temp=i;while(temp<xLength&&board[temp][j]=='X'){isGoneThroguh[temp][j]=1;temp++;}count++;}else{count++;}}}}return count;}public static void main(String[] args) {// TODO Auto-generated method stubBattleships_in_a_Board_419 b = new Battleships_in_a_Board_419();char[][] a=new char[][]{{'X','.','.','X'},{'.','.','.','X'},{'.','.','.','X'}};System.out.println(b.countBattleships(a));}}
然后大神的思路令我膜拜:Going over all cells, we can count only those that are the "first" cell of the battleship. First cell will be defined as the most top-left cell. We can check for first cells by only counting cells that do not have an 'X' to the left and do not have an 'X' above them.
只需要数舰队中最上左的那只舰就好了。即你发现了一个'X'之后,如果它上面和左边没'X'的话,它就是最上左的舰,把count+1,反之则不是最上左的舰,直接跳过。
public int countBattleships(char[][] board) { int m = board.length; if (m==0) return 0; int n = board[0].length; int count=0; for (int i=0; i<m; i++) { for (int j=0; j<n; j++) { if (board[i][j] == '.') continue; if (i > 0 && board[i-1][j] == 'X') continue; if (j > 0 && board[i][j-1] == 'X') continue; count++; } } return count; }大神的解法多么简洁,多么漂亮。。。这真是智商的差别啊T_T!!
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