LeetCode題解：Battleships in a Board

2019-11-14 10:25:56

字體：大中小

供稿：網(wǎng)友

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...XIn the above board there are 2 battleships.Invalid Example:
...XXXXX...XThis is an invalid board that you will not receive - as battleships will always have a cell separating between them.
思路：
簡(jiǎn)單的思路是做搜索。但是考慮到每個(gè)船和每個(gè)船之間至少橫向縱向有一個(gè)空格，那么如果只考慮每艘船的右下角：
XX.X...X
這樣的話，右下角的X右邊和下面都是空格，每次碰到符合這個(gè)條件的X就認(rèn)為碰到一個(gè)船，否則不管。
題解：
int countBattleships(const std::vector<std::vector<char>>& board) {    const int M = board.size();    const int N = board[0].size();    int numShips(0);    for(int i = 0; i < M; ++i) {        for(int j = 0; j < N; ++j) {            if (board[i][j] == 'X') {                numShips += ((i < M - 1 && board[i + 1][j] == '.') || (i == M - 1)) &&                            ((j < N - 1 && board[i][j + 1] == '.') || (j == N - 1));            }        }    }    return numShips;}