On a 3x3 board, there are 8 tiles represented by the integers 1 through 8, and an empty square represented by 0.
A move consists of choosing 0 and a 4-directionally adjacent number and swapping it.
Given an initial state of the puzzle board and final state, return the least number of moves required so that the initial state to final state.
If it is impossible to move from initial state to final state, return -1.
Example 1:
Input:
[
[2,8,3],
[1,0,4],
[7,6,5]
]
[
[1,2,3],
[8,0,4],
[7,6,5]
]
Output:
4
Explanation:
[ [
[2,8,3], [2,0,3],
[1,0,4], --> [1,8,4],
[7,6,5] [7,6,5]
] ]
[ [
[2,0,3], [0,2,3],
[1,8,4], --> [1,8,4],
[7,6,5] [7,6,5]
] ]
[ [
[0,2,3], [1,2,3],
[1,8,4], --> [0,8,4],
[7,6,5] [7,6,5]
] ]
[ [
[1,2,3], [1,2,3],
[0,8,4], --> [8,0,4],
[7,6,5] [7,6,5]
] ]
Example 2：
Input: [[2,3,8],[7,0,5],[1,6,4]] [[1,2,3],[8,0,4],[7,6,5]] Output: -1
public class Solution { /** * @param init_state: the initial state of chessboard * @param final_state: the final state of chessboard * @return: return an integer, denote the number of minimum moving */ public int minMoveStep(int[][] init_state, int[][] final_state) { // # write your code here String start = matrixToStr(init_state); String end = matrixToStr(final_state); int step = 0; Set<String> visited = new HashSet<>(); Queue<String> queue = new LinkedList<>(); queue.offer(start); visited.add(start); while (!queue.isEmpty()) { int size = queue.size(); while (size-- > 0) { String cur = queue.poll(); if (cur.equals(end)) { return step; } for (String nxt: getNext(cur)) { if (visited.contains(nxt)) { continue; } queue.offer(nxt); visited.add(nxt); } } step += 1; } return -1; } private String matrixToStr(int[][] matrix) { StringBuilder sb = new StringBuilder(); for (int i = 0; i < matrix.length; i++) { for (int j = 0; j < matrix[0].length; j++) { sb.append(matrix[i][j]); } } return sb.toString(); } private List<String> getNext(String cur) { List<String> list = new ArrayList<>(); int zeroIdx = cur.indexOf("0"); int curRow = zeroIdx / 3; int curCol = zeroIdx % 3; int[][] directions = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}}; for (int[] direction : directions) { int nxtRow = curRow + direction[0]; int nxtCol = curCol + direction[1]; if (nxtRow < 0 || nxtRow >= 3 || nxtCol < 0 || nxtCol >= 3) { continue; } char[] charArr = cur.toCharArray(); charArr[zeroIdx] = charArr[nxtRow * 3 + nxtCol]; charArr[nxtRow * 3 + nxtCol] = ‘0‘; list.add(new String(charArr)); } return list; } }
[LintCode] 794. Sliding Puzzle II
原文：https://www.cnblogs.com/xuanlu/p/12571964.html