419. Battleships in a Board

最新推荐文章于 2019-11-15 10:14:32 发布

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本文介绍了一种在一维数组中高效计数战舰数量的方法。通过遍历二维矩阵，利用状态标志判断战舰的出现与结束，实现了一次遍历且仅使用常数额外空间的目标。文章提供了一个具体的示例和代码实现。

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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) 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?

#include <iostream>
#include <cstdio>
#include <string>
#include <vector>
const int maxn = 1000;
using namespace std;

 class Solution {
    public:
        int countBattleships(vector< vector<char> > &board) {
            int n = board.size();
            int m = board[0].size();
            vector< vector<char> > row = board;
            int cnt = 0;
            bool flag = false;
            for(int i = 0; i < n; i++) {
                for(int j = 0; j < m; j++) {
                        if(i == 0) { //第一行情况
                           if(j == 0 && row[0][0] == 'X') {
                                  flag = true;
                                  cnt++;
                           } else {
                                if(flag == false && row[i][j] == 'X') {
                                        flag = true;
                                        cnt++;
                                }
                                else if(flag == true && row[i][j] == '.') {
                                        flag = false;
                                }
                           }
                        } else {//非第一行的情况
                                if(j == 0) {
                                        if(row[i-1][0] == 'X' && row[i][0] == 'X') {
                                                flag = true;
                                        } else if(row[i-1][0] == 'X' && row[i][0] == '.') {
                                                flag = false;
                                        } else if(row[i-1][0] == '.' && row[i][0] == 'X') {
                                                flag = true;
                                                cnt++;
                                        } else if(row[i-1][0] == '.' && row[i][0] == '.') {
                                                flag = false;
                                        }

                                } else {
                                        if(flag == false && row[i][j] == 'X' && row[i-1][j] == '.') {
                                                flag = true;
                                                cnt++;
                                        } else if(flag == false && row[i][j] == 'X' && row[i-1][j] == 'X') {
                                                flag = true;
                                        } else if(flag == true && row[i][j] == '.') {
                                                flag = false;
                                        }
                                }

                        }
                }
            }
            return cnt;
        }
};

int main()
{
    vector<char> row[maxn];
    vector< vector<char> > graph;
    int n, m;
    while(scanf("%d%d", &n, &m) != EOF) {
        char ch;
        getchar();
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                scanf("%c", &ch);
                row[i].push_back(ch);
            }
            getchar();
            graph.push_back(row[i]);
        }
        Solution *obj = new Solution();
        int result =  obj->countBattleships(graph);
        cout << result << endl;
    }
    return 0;
}
/**
3 4
X..X
...X
...X

3 4
...X
XX.X
...X

4 8
..XXX.XX
X....X..
X.XX.X.X
.X..X..X

**/