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:
...X XXXX ...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?
给定一个数组,X表示船 .表示空,求一共有多少条船。
其中:如果一个X的纵向或者横向上有X,则这俩是一条船。以此类推。
这道题有个附属条件是只用复杂度为 O(1)的额外内存并且不能修改入参。
思路:一个X的左边或者上面如果没有X,则船数+1;
/** * @param {character[][]} board * @return {number} */ var countBattleships = function(board) { var count = 0; for(var i=0;i<board.length;i++){ for(var j=0;j<board[i].length;j++){ if(i>0&&j>0){ if(board[i-1][j]==='.'&&board[i][j-1]==='.'&&board[i][j]==="X"){ count++; } }else if(i>0&&j===0){ if(board[i-1][j]==='.'&&board[i][j]==="X"){ count++; } }else if(j>0&&i===0){ if(board[i][j-1]==='.'&&board[i][j]==="X"){ count++; } }else{ //(i===0&&j===0)===true if(board[i][j]==="X"){ count++; } } } } return count; };