LeetCode.419.Battleships in a Board 甲板上的战舰

题目描述

419 Battleships in a Board
https://leetcode-cn.com/problems/battleships-in-a-board/
https://leetcode.com/problems/battleships-in-a-board/description/

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?

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解题过程

这道题刚看的时候没思路，感觉挺复杂挺乱的，然后多读几遍题意，用测试用例试着分析分析，慢慢就有思路了。
题意就是有一个二维矩阵，计算里面连续的数列(1xN,Nx1)个数，数列只能是水平或垂直连续的，不会斜着出现，并且也不会有任意两个数列挨着。
解题思路最基本的是遍历二维矩阵遇到x就加1，但要排除连续数列的重复加1，什么情况下有连续数列呢？按从左到右、从上到下遍历来说，只能是当前位置左边或上边挨着已经有x的情况，所以再把排除情况考虑进去即可。

代码如下，满足一次遍历、O(1)额外空间、不修改原数组的进阶要求。但是效率不高，我多了一步，先加再减费时了，看别人提交的高效代码，思路是一样的，但稍微处理下能进一步缩短时间。

private static class SolutionV2018 {
    public int countBattleships(char[][] board) {
        int count = 0;
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[i].length; j++) {
                // 新出现x
                if (board[i][j] == 'X') {
                    count++; //先加1
                    // 如果左边是x，不能重复计数，减1
                    if (j >= 1 && board[i][j - 1] == 'X') {
                        count--;
                    }
                    // 如果上边是x，不能重复计数，减1
                    if (i >= 1 && board[i - 1][j] == 'X') {
                        count--;
                    }
                }
            }
        }
        return count;
    }
}

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GitHub代码

algorithms/leetcode/leetcode/_419_BattleshipsInBoard.java
https://github.com/masikkk/algorithms/blob/master/leetcode/leetcode/_419_BattleshipsInBoard.java

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