甲板上的战舰（Battleships in a Board）

419. Battleships in a Board

来源： LeetCode 419. Battleships in a Board

题目描述

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?

思路分析

BFS思路改变原矩阵

代码

class Solution {
# define REP(i, a) for(int i=0; i<(a); ++i)
public:
    int dx[4] = {0,0,1,-1}, dy[4] = {1,-1,0,0};
    int countBattleships(vector<vector<char>>& board) {
        // four toward BFS
        if(!board.size() || !board[0].size()) return 0;
        int res = 0; 
        REP(i, board.size()){
            REP(j, board[i].size()){
                if(board[i][j] == 'X'){
                    board[i][j] = '.';
                    bfs(board, i, j);
                    ++res;
                }
            }
        }
        return res;
    }
    void bfs(vector<vector<char>>& board, int x, int y){
        REP(i, 4){
            x += dx[i]; y += dy[i];
            if(x>=0 && x<board.size() && y>= 0 && y<board[0].size() && board[x][y] == 'X'){
                board[x][y] = '.';
                bfs(board, x, y);
            }
            x -= dx[i]; y -= dy[i];
        }
    }
};

代码改进

进阶一次扫描，不改变原数组

class Solution {
# define REP(i, a) for(int i=0; i<(a); ++i)
public:
    int countBattleships(vector<vector<char>>& board) {
        // 螺旋遍历，舰与舰之间至少有空格，左上角向右下角扫描
        // 改变判断
        if(!board.size() || !board[0].size()) return 0;
        int res = 0;
        REP(i, board.size()){
            REP(j, board[0].size()){
                if(i >0 && j > 0){
                    if(board[i][j] == 'X' && board[i-1][j] != 'X' && board[i][j-1] != 'X') ++res;
                }
                else if(board[i][j] == 'X'){
                    if(i > 0 && board[i-1][j] == '.') ++res;
                    else if(j > 0 && board[i][j-1] == '.') ++res;
                    else if(i==0 && j == 0) ++res;
                } // i==0 || j==0 || i ==0 && j == 0
            }
        }
        return res;
    }
};