LeetCode:419. Battleships in a Board

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?

题意：给定字符矩阵，其满足以下条件：

1）只存在“X” 和 ”.“。

2）X只按照行或者列进行排列

3）行列不相交。

计算X整体的个数。

分析：

算法一：DFS

找到一个未标记的X，然后利用DFS将与之连接的X全部标记，并且计数增1，然后找到下一个未标记的X，依次利用DFS标记，直到全部标记。

这个算法适用于任意形状的X的计数问题。

算法空间复杂度O(n^2)，时间复杂度则为O(n^2)（邻接矩阵表示O(n^2)，如果是邻接表表示则时间复杂度是O(V+e)）。

代码如下：

class Solution {
public int countBattleships(char[][] board) {
if(board==null)
return 0;
int n=board.length;
if(n==0)
return 0;
int m=board[0].length;
int count=0;
boolean[][]flag=new boolean
[m];
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
if(board[i][j]=='X'&&!flag[i][j]){
dfs(board,flag,i,j,n,m);
count++;
}
}
}
return count;
}
public void dfs(char[][]board,boolean[][]flag,int i,int j,int n,int m){
flag[i][j]=true;
if(i-1>=0&&board[i-1][j]=='X'&&!flag[i-1][j])
dfs(board,flag,i-1,j,n,m);
if(j-1>=0&&board[i][j-1]=='X'&&!flag[i][j-1])
dfs(board,flag,i,j-1,n,m);
if(i+1<n&&board[i+1][j]=='X'&&!flag[i+1][j])
dfs(board,flag,i+1,j,n,m);
if(j+1<m&&board[i][j+1]=='X'&&!flag[i][j+1])
dfs(board,flag,i,j+1,n,m);
}
}

算法二：

利用题目中的信息2，可以构造专门针对此题的算法：

按行扫描，如果当前节点是”X”，则判断其左边（行）和上边（列）是否仍然是”X”，如果都不是，则计数加一。

算法满足空间复杂度O(1)，时间复杂度O(n^2)，并且是一遍扫描。

代码如下：

class Solution {
public int countBattleships(char[][] board) {
if(board==null)
return 0;
int n=board.length;
if(n==0)
return 0;
int m=board[0].length;
int count=0;
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
if(board[i][j]=='X'){
boolean flag=false;
if(i-1>=0&&board[i-1][j]=='X')
flag=true;
if(j-1>=0&&board[i][j-1]=='X')
flag=true;
if(!flag)
count++;
}
}
}
return count;
}
}