https://leetcode.com/problems/battleships-in-a-board
Given an 2D board, count how many different battleships are in it. The battleships are represented with
X.
http://bookshadow.com/weblog/2016/10/13/leetcode-battleships-in-a-board/
这道题让我们求战舰的个数，所谓的战舰就是只能是一行或者一列，不能有拐弯。这道题降低了难度的做法是限定了不会有相邻的两个战舰的存在，有了这一点限制，那么我们只需要遍历一次二维数组就行了，只要找出战舰的起始点。所谓的战舰起始点，就是为X的点，而且该点的上方和左边的点不能为X，所以我们只要找出所有满足这个条件的点即可
https://leetcode.com/problems/battleships-in-a-board/discuss/90940/4ms-Java-Optimized-code
https://discuss.leetcode.com/topic/65418/o-n-2-time-and-o-1-space-without-modifying-the-board
http://bookshadow.com/weblog/2016/10/13/leetcode-battleships-in-a-board/
X.
https://discuss.leetcode.com/topic/63940/java-dfs-solution-6ms
X. https://leetcode.com/problems/battleships-in-a-board/discuss/90957/DFS-and-BFS-Flood-Fill-Algorithm-with-C%2B%2B
Given an 2D board, count how many different 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?http://bookshadow.com/weblog/2016/10/13/leetcode-battleships-in-a-board/
由于board中的战舰之间确保有'.'隔开，因此遍历board，若某单元格为'X'，只需判断其左边和上边的相邻单元格是否也是'X'。
如果左邻居或者上邻居单元格是'X'，则说明当前单元格是左边或者上边战舰的一部分；
否则，令计数器+1
http://www.voidcn.com/article/p-xncmffwh-bqs.html
Because the board is always valid which means at least one horizontal or vertical cell separates between two battleships. We can check for the first cells by only counting cells that do not have an 'X' to the left and do not have an 'X' above them.
http://www.cnblogs.com/grandyang/p/5979207.html这道题让我们求战舰的个数，所谓的战舰就是只能是一行或者一列，不能有拐弯。这道题降低了难度的做法是限定了不会有相邻的两个战舰的存在，有了这一点限制，那么我们只需要遍历一次二维数组就行了，只要找出战舰的起始点。所谓的战舰起始点，就是为X的点，而且该点的上方和左边的点不能为X，所以我们只要找出所有满足这个条件的点即可
https://leetcode.com/problems/battleships-in-a-board/discuss/90940/4ms-Java-Optimized-code
Aheadof a battleship means thetop mostorleft mostcell with value 'X'.
Thus, we only need to count thoseheads.
There are three rules to tell if a cell is a 'head':
- The cell is a 'X' (
board[i][j] == 'X') - No left side neighbor, or the left neighbor is a '.' (
i == 0 || board[i - 1][j] == '.') - No right side neighbor, or the right neighbor is a '.' (
j == 0 || board[i][j - 1] == '.')
public int countBattleships(char[][] board) {
if (board == null || board.length == 0 || board[0].length == 0) return 0;
int R = board.length, C = board[0].length, cnt = 0;
for (int i = 0; i < R; i++) {
for (int j = 0; j < C; j++) {
if (board[i][j] == 'X' && (i == 0 || board[i - 1][j] == '.') && (j == 0 || board[i][j - 1] == '.'))
cnt++;
}
}
return cnt;
}
...Note...
For a statement like
So, for the best performance, we write the program in this way, instead of using a lot of if statements.
https://leetcode.com/problems/battleships-in-a-board/discuss/90902/Simple-Java-SolutionFor a statement like
if (A && B && C), when A is false, the program will not compute B and C.So, for the best performance, we write the program in this way, instead of using a lot of if statements.
Going over all cells, marking 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.
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;
}https://discuss.leetcode.com/topic/65418/o-n-2-time-and-o-1-space-without-modifying-the-board
public int countBattleships(char[][] board) {
int count = 0;
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
if (board[i][j] == 'X' && (i == 0 || board[i-1][j] == '.') && (j == 0 || board[i][j-1] == '.')) {
count++;
}
}
}
return count;
}Aheadof a battleship means thetop mostorleft mostcell with value 'X'.
Thus, we only need to count thoseheads.
There are three rules to tell if a cell is a 'head':
- The cell is a 'X' (
board[i][j] == 'X') - No left side neighbor, or the left neighbor is a '.' (
i == 0 || board[i - 1][j] == '.') - No right side neighbor, or the right neighbor is a '.' (
j == 0 || board[i][j - 1] == '.')
Code:
public int countBattleships(char[][] board) {
if (board == null || board.length == 0 || board[0].length == 0) return 0;
int R = board.length, C = board[0].length, cnt = 0;
for (int i = 0; i < R; i++) {
for (int j = 0; j < C; j++) {
if (board[i][j] == 'X' && (i == 0 || board[i - 1][j] == '.') && (j == 0 || board[i][j - 1] == '.'))
cnt++;
}
}
return cnt;
}
...Note...
For a statement like
So, for the best performance, we write the program in this way, instead of using a lot of if statements.
For a statement like
if (A && B && C), when A is false, the program will not compute B and C.So, for the best performance, we write the program in this way, instead of using a lot of if statements.
http://bookshadow.com/weblog/2016/10/13/leetcode-battleships-in-a-board/
由于board中的战舰之间确保有'.'隔开，因此遍历board，若某单元格为'X'，只需判断其左边和上边的相邻单元格是否也是'X'。
如果左邻居或者上邻居单元格是'X'，则说明当前单元格是左边或者上边战舰的一部分；
否则，令计数器+1
X.
https://discuss.leetcode.com/topic/63940/java-dfs-solution-6ms
Basic idea: If any 'X' has more than one neighbors is 'X' then these connected 'X' cannot form a valid Battleship.
public int countBattleships(char[][] board) {
int M = board.length;
int N = board[0].length;
boolean[][] marked = new boolean[M][N];
int cnt = 0;
for (int i = 0; i < M; i++) {
for (int j = 0; j < N; j++) {
if (board[i][j] == 'X' && !marked[i][j] && dfs(board, i, j, marked)) cnt++;
}
}
return cnt;
}
public boolean dfs(char[][] board, int r, int c, boolean[][] marked) {
marked[r][c] = true;
int[] direct = {1, 0, -1, 0, 1};
int cnt = 0;
boolean res = true;
for (int i = 0; i < 4; i++) {
int newR = r + direct[i];
int newC = c + direct[i + 1];
if (newR >= 0 && newR < board.length && newC >= 0 && newC < board[0].length && !marked[newR][newC] && board[newR][newC] == 'X') {
cnt++;
if (!dfs(board, newR, newC, marked)) res = false;
}
}
return cnt > 1 || res;
}
解法II FloodFill
遍历board，用DFS（深度优先搜索）对每一个'X'位置进行探索与标记，同时进行计数。
def countBattleships(self, board):
"""
:type board: List[List[str]]
:rtype: int
"""
vs = set()
h = len(board)
w = len(board[0]) if h else 0
def dfs(x, y):
for dx, dy in zip((1, 0, -1, 0), (0, 1, 0, -1)):
nx, ny = x + dx, y + dy
if 0 <= nx < h and 0 <= ny < w:
if (nx, ny) not in vs and board[nx][ny] == 'X':
vs.add((nx, ny))
dfs(nx, ny)
ans = 0
for x in range(h):
for y in range(w):
if (x, y) not in vs and board[x][y] == 'X':
ans += 1
vs.add((x, y))
dfs(x, y)
return ansX. https://leetcode.com/problems/battleships-in-a-board/discuss/90957/DFS-and-BFS-Flood-Fill-Algorithm-with-C%2B%2B
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;
}