标签:ems [] not complex sha problems ships sep log
原题链接在这里:https://leetcode.com/problems/battleships-in-a-board/#/description
题目:
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:
1xN (1 row, N columns) or Nx1 (N rows, 1 column), where N can be of any size.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?
题解:
无论是1*N 还是 N*1型的battleship, 都以左上角的X来标志. 只有遇到X, 并且左边或者上边没有连着的X时才算新的battleship.
Time Complexity: O(mn), m = board.length, n = board[0].length.
Space: O(1).
AC Java:
1 public class Solution { 2 public int countBattleships(char[][] board) { 3 if(board == null || board.length == 0 || board[0].length == 0){ 4 return 0; 5 } 6 7 int res = 0; 8 for(int i = 0; i<board.length; i++){ 9 for(int j = 0; j<board[0].length; j++){ 10 if(board[i][j] == ‘.‘){ 11 continue; 12 } 13 if(i>0 && board[i-1][j]==‘X‘){ 14 continue; 15 } 16 if(j>0 && board[i][j-1]==‘X‘){ 17 continue; 18 } 19 res++; 20 } 21 } 22 return res; 23 } 24 }
LeetCode Battleships in a Board
标签:ems [] not complex sha problems ships sep log
原文地址:http://www.cnblogs.com/Dylan-Java-NYC/p/6751562.html