LWC 69: 773. Sliding Puzzle
Problem:
On a 2x3 board, there are 5 tiles represented by the integers 1 through 5, and an empty square represented by 0.
A move consists of choosing 0 and a 4-directionally adjacent number and swapping it.
The state of the board is solved if and only if the board is [[1,2,3],[4,5,0]].
Given a puzzle board, return the least number of moves required so that the state of the board is solved. If it is impossible for the state of the board to be solved, return -1.
Examples:
Input: board = [[1,2,3],[4,0,5]]
Output: 1
Explanation: Swap the 0 and the 5 in one move.
Input: board = [[1,2,3],[5,4,0]]
Output: -1
Explanation: No number of moves will make the board solved.
Input: board = [[4,1,2],[5,0,3]]
Output: 5
Explanation: 5 is the smallest number of moves that solves the board.
An example path:
After move 0: [[4,1,2],[5,0,3]]
After move 1: [[4,1,2],[0,5,3]]
After move 2: [[0,1,2],[4,5,3]]
After move 3: [[1,0,2],[4,5,3]]
After move 4: [[1,2,0],[4,5,3]]
After move 5: [[1,2,3],[4,5,0]]
Input: board = [[3,2,4],[1,5,0]]
Output: 14
Note:
- board will be a 2 x 3 array as described above.
- board[i][j] will be a permutation of [0, 1, 2, 3, 4, 5].
思路：
狀態數很少，BFS套路題，把[[1, 2, 3],[4, 5, 0]]hash成字符串”123450”即可。
Java版本：
public int slidingPuzzle(int[][] board) {
int n = board.length;
int m = board[0].length;
String ter = "123450";
Queue<int[][]> q = new ArrayDeque<>();
Set<String> vis = new HashSet<>();
q.offer(board);
vis.add(hash(board));
int turn = 0;
int[][] dir = {{1, 0},{-1, 0},{0, 1},{0, -1}};
while (!q.isEmpty()) {
int size = q.size();
for (int i = 0; i < size; ++i) {
int[][] now = q.poll();
if (hash(now).equals(ter)) {
return turn;
}
// find 0
int x = -1;
int y = -1;
for (int k = 0; k < n; ++k) {
for (int l = 0; l < m; ++l) {
if (now[k][l] == 0) {
x = k;
y = l;
break;
}
}
}
for (int[] d : dir) {
int nx = d[0] + x;
int ny = d[1] + y;
if (nx >= 0 && nx < n && ny >= 0 && ny < m) {
int[][] aux = clone(now);
swap(aux, x, y, nx, ny);
if (!vis.contains(hash(aux))) {
q.offer(aux);
vis.add(hash(aux));
}
}
}
}
turn ++;
}
return -1;
}
int[][] clone(int[][] board) {
int n = board.length;
int m = board[0].length;
int[][] aux = new int[n][m];
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
aux[i][j] = board[i][j];
}
}
return aux;
}
String hash(int[][] board) {
StringBuilder sb = new StringBuilder();
int n = board.length;
int m = board[0].length;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
sb.append(board[i][j]);
}
}
return sb.toString();
}
void swap(int[][] board, int ai, int aj, int bi, int bj) {
int tmp = board[ai][aj];
board[ai][aj] = board[bi][bj];
board[bi][bj] = tmp;
}Python版本：
class Solution(object):
def slidingPuzzle(self, board):
"""
:type board: List[List[int]]
:rtype: int
"""
seen = set()
END = '123450'
INIT = ''.join(map(str, board[0])) + ''.join(map(str, board[1]))
queue = []
seen.add(INIT)
queue.append(INIT)
turn = 0
while queue:
size = len(queue)
for i in range(size):
now = queue.pop(0)
if now == END: return turn
idx = now.index('0')
r = idx / 3
c = idx % 3
for d in (1, -1):
for nx, ny in ((r + d, c), (r, c + d)):
if nx >= 0 and nx < 2 and ny >= 0 and ny < 3:
nidx = nx * 3 + ny
nxt = list(now)
nxt[idx], nxt[nidx] = nxt[nidx], nxt[idx]
nxt = ''.join(nxt)
if nxt not in seen:
seen.add(nxt)
queue.append(nxt)
turn += 1
return -1 
