标签：htm http color site 参考 win size 空间 lan
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:
1xN (1 row, N columns) or Nx1 (N rows, 1 column), where N can be of any size.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?
思路：
首先想到dfs。
void dfsbattle(vector<vector<char>>& board,vector<vector<bool>>& visited,int i,int j) { int m = board.size(), n = board[0].size(); if (i < 0 || i >= m || j < 0 || j >= n || visited[i][j] || board[i][j] == ‘.‘)return; visited[i][j] = true; dfsbattle(board, visited, i + 1, j); dfsbattle(board, visited, i - 1, j); dfsbattle(board, visited, i , j+1); dfsbattle(board, visited, i , j-1); } int countBattleships(vector<vector<char>>& board) { if (board.empty())return 0; int m = board.size(), n = board[0].size(); vector<vector<bool>>visited(m, vector<bool>(n, false)); int ret = 0; for (int i = 0; i < m;i++) { for (int j = 0; j < n;j++) { if (board[i][j]==‘X‘&& !visited[i][j]) { dfsbattle(board, visited, i,j); ret++; } } } return ret; }
又题目提到不用额外空间，而且只遍历一次。。
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]!=‘X‘) && (j==0 || board[i][j-1]!=‘X‘)) count++; return count; }
参考：
https://discuss.leetcode.com/topic/64027/share-my-7-line-code-1-line-core-code-3ms-super-easy
[leetcode-419-Battleships in a Board]
标签：htm http color site 参考 win size 空间 lan
原文地址：http://www.cnblogs.com/hellowooorld/p/7152990.html
