[leetcode]419. Battleships in a Board

[leetcode]419. Battleships in a Board

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Analysis

周四不知道快不快乐—— [有点想回家过暑假的冲动！]

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.

Implement

方法一（直接遍历）

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

方法二（DFS）

class Solution {
public:
    int countBattleships(vector<vector<char>>& board) {
        row = board.size();
        col = board[0].size();
        visit.resize(row);
        for(int i=0; i<row; i++)
            visit[i].resize(col, false);
        int res = 0;
        for(int i=0; i<row; i++){
            for(int j=0; j<col; j++){
                if(!visit[i][j] && board[i][j] == 'X'){
                    res++;
                    DFS(i, j, board);
                }
            }
        }
        return res;
    }
    void DFS(int i, int j, vector<vector<char>>& board){
        visit[i][j] = true;
        for(int k=0; k<4; k++){
            if(i+di[k]>=0 && i+di[k]<row && j+dj[k]>=0 && j+dj[k]<col && !visit[i+di[k]][j+dj[k]] && board[i+di[k]][j+dj[k]]=='X'){
                visit[i+di[k]][j+dj[k]] = true;
                DFS(i+di[k], j+dj[k], board);
            }
        }
    }
private:
    vector<vector<bool>> visit;
    int di[4] = {1, -1, 0, 0};
    int dj[4] = {0, 0, 1, -1};
    int row, col;
};