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Java实现 LeetCode 773 滑动谜题(BFS)

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773. 滑动谜题

在一个 2 x 3 的板上(board)有 5 块砖瓦,用数字 1~5 来表示, 以及一块空缺用 0 来表示.

一次移动定义为选择 0 与一个相邻的数字(上下左右)进行交换.

最终当板 board 的结果是 [[1,2,3],[4,5,0]] 谜板被解开。

给出一个谜板的初始状态,返回最少可以通过多少次移动解开谜板,如果不能解开谜板,则返回 -1 。

示例:

输入:board = [[1,2,3],[4,0,5]]
输出:1
解释:交换 0 和 5 ,1 步完成
输入:board = [[1,2,3],[5,4,0]]
输出:-1
解释:没有办法完成谜板
输入:board = [[4,1,2],[5,0,3]]
输出:5
解释:
最少完成谜板的最少移动次数是 5 ,
一种移动路径:
尚未移动: [[4,1,2],[5,0,3]]
移动 1 次: [[4,1,2],[0,5,3]]
移动 2 次: [[0,1,2],[4,5,3]]
移动 3 次: [[1,0,2],[4,5,3]]
移动 4 次: [[1,2,0],[4,5,3]]
移动 5 次: [[1,2,3],[4,5,0]]
输入:board = [[3,2,4],[1,5,0]]
输出:14
提示:

board 是一个如上所述的 2 x 3 的数组.
board[i][j] 是一个 [0, 1, 2, 3, 4, 5] 的排列.

PS:
BFS搜索,交换,然后每一次都看是不是匹配

class Solution {
    private static int ROW = 2;
    private static int COL = 3;
    private static final String RESULT = "123450";
    Set<String> set;
    LinkedList<String> queue;
    
    public int slidingPuzzle(int[][] board) {
    	if ((board.length != ROW) || (board[0].length != COL)) {
    		return 0;
    	}
    	set = new HashSet<>();
    	int time = 0;
    	char[] boardStr = getCharFromBoard(board);
    	if (RESULT.equals(Arrays.toString(boardStr))) {
    		return 0;
    	}
    	queue = new LinkedList<>();
    	queue.addLast(new String(boardStr));
    	while (!queue.isEmpty()) {
    		int size = queue.size();
    		for (int i = 0; i < size; i++) {
    			String node = queue.pollFirst();
    			if (RESULT.equals(node)) {
    				return time;
    			}
    			boardStr = node.toCharArray();
    			if (boardStr[0] == '0') {
    				move(boardStr, 0, 1);
    				move(boardStr, 0, 3);
    			} else if (boardStr[1] == '0') {
    				move(boardStr, 1, 0);
    				move(boardStr, 1, 2);
    				move(boardStr, 1, 4);
    			} else if (boardStr[2] == '0') {
    				move(boardStr, 2, 1);
    				move(boardStr, 2, 5);
    			} else if (boardStr[3] == '0') {
    				move(boardStr, 3, 0);
    				move(boardStr, 3, 4);
    			} else if (boardStr[4] == '0') {
    				move(boardStr, 4, 1);
    				move(boardStr, 4, 3);
    				move(boardStr, 4, 5);
    			} else if (boardStr[5] == '0') {
    				move(boardStr, 5, 2);
    				move(boardStr, 5, 4);
    			}
    		}
    		time++;
    	}
    	return -1;
    }
    
    private void move(char[] string, int i, int j) {
		swap(string, i, j);
		String str = new String(string);
		if (!set.contains(str)) {
			queue.addLast(str);
			set.add(str);
		}
		swap(string, i, j);
    }
    
    private void swap (char[] string, int i, int j) {
    	char temp = string[i];
    	string[i] = string[j];
    	string[j] = temp;
    }
    
    private char[] getCharFromBoard(int[][] board) {
    	char[] res = new char[6];
    	res[0] = (char) ('0' + board[0][0]);
    	res[1] = (char) ('0' + board[0][1]);
    	res[2] = (char) ('0' + board[0][2]);
    	res[3] = (char) ('0' + board[1][0]);
    	res[4] = (char) ('0' + board[1][1]);
    	res[5] = (char) ('0' + board[1][2]);
    	return res;
    }
}
   

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来源: https://blog.51cto.com/u_14827074/3032170