【LeetCode】419.Battleships in a Board(Medium)解题报告
2017-11-05 16:42
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【LeetCode】419.Battleships in a Board(Medium)解题报告
题目地址:https://leetcode.com/problems/battleships-in-a-board/description/
题目描述:
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:
1. You receive a valid board, made of only battleships or empty slots.
2. 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.
3. At least one horizontal or vertical cell separates between two battleships - there are no adjacent battleships.
Example:
In the above board there are 2 battleships.
Invaild Example:
This is an invalid board that you will not receive - as battleships will always have a cell separating between them.
Solutions:
正解:
其实思想很简单,但是开始理解错了,以为不合理例子要自己排除,注意下X是大写的就行了。
Date:2017年11月5日
题目地址:https://leetcode.com/problems/battleships-in-a-board/description/
题目描述:
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:
1. You receive a valid board, made of only battleships or empty slots.
2. 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.
3. 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.
Invaild Example:
...X XXXX ...X
This is an invalid board that you will not receive - as battleships will always have a cell separating between them.
Solutions:
正解:
class Solution { public int countBattleships(char[][] board) { if(board.length==0 || board[0].length==0){ return 0; } int count=0; for(int i=0 ; i<board.length ;i++){ for(int j=0 ;j<board[0].length ; j++){ if(board[i][j]=='X'){ if(i>0 && board[i][j] == board[i-1][j]){ continue; } if(j>0 && board[i][j] == board[i][j-1]){ continue; } count++; } } } return count; } }
其实思想很简单,但是开始理解错了,以为不合理例子要自己排除,注意下X是大写的就行了。
Date:2017年11月5日
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