Battleships in a Board战列舰队问题解法分析

问题详见：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?

解题思路：

由题目可知该问题中战列舰队是有在同一行或者同一列的X战舰组成的舰队，而且两个战列舰队之间不会重复和毗邻。所以我们可以用DFS和BFS的方法对其进行查找。下面是两种算法的C++代码和提交结果图。

DFS：

class Solution {
public:
int m, n;
vector<vector<bool>> flag;
int go[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

void dfs(vector<vector<char>>& board, int i, int j) {
if (i < 0 || i >= m || j < 0 || j >= n || board[i][j] == '.' || flag[i][j]) return;
flag[i][j] = true;
for (int d = 0; d < 4; ++d) dfs(board, i+go[d][0], j+go[d][1]);
}

int countBattleships(vector<vector<char>>& board) {
if (board.empty()) return 0;
m = board.size(), n = board[0].size();
flag.resize(m, vector<bool>(n, false));
int result = 0;
for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j)
if (board[i][j] == 'X' && !flag[i][j]) {
++result;
dfs(board, i, j);
}
return result;
}
};

BFS：

class Solution {
public:
int go[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
int countBattleships(vector<vector<char>>& board) {
if (board.empty()) return 0;
int m = board.size(), n = board[0].size();
vector<vector<bool>> flag(m, vector<bool>(n, false));
int result = 0;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (board[i][j] == 'X' && !flag[i][j]) {
++result;

queue<pair<int, int>> q;
q.push({i, j});
while (!q.empty()) {
auto t = q.front(); q.pop();
flag[t.first][t.second] = true;
for (int d = 0; d < 4; ++d) {
int ni = t.first+go[d][0], nj = t.second+go[d][1];
if (ni < 0 || ni >= m || nj < 0 || nj >= n || board[ni][nj] == '.' || flag[ni][nj]) continue;
q.push({ni, nj});
}
}
}
}
}
return result;
}
};

然而我们发现还有一种更加简单的方法可以得出结果。那就是根据题目中对战列舰队的严格限制的规则的利用。我们可以判断搜索到的点是属于已经查找到的战列舰队还是一个新的战列舰队中的一个战舰，只需要判断这个点在2D空间的上一个和左一个位置是否存在战舰（搜索按照从上往下从左到右的顺序），如果存在说明该战舰属于已经查找到的战列舰队，否则就是新的战列舰队（根据题意即使只有一个战舰也能构成战列舰队）。所以我们得出如下的简单算法：

class Solution {
public:
int countBattleships(vector<vector<char>>& board) {
int count = 0;
for(int i=0;i<board.size();i++)
for(int j=0;j<board[0].size();j++)
if(board[i][j]=='X' && (i==0 || board[i-1][j]!='X') && (j==0 || board[i][j-1]!='X')) count++;
return count;
}
};

提交运行结果：