【LeetCode】419. Battleships in a Board

最新推荐文章于 2019-11-15 10:14:32 发布

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本文介绍了一种用于计数二维棋盘上战舰数量的算法。战舰由字符 'X' 表示，且只能水平或垂直放置。通过检查每个 'X' 的左侧和上方是否有 'X' 来判断是否为新战舰。

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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) 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?

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;
};