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: 1Explanation: Swap the 0 and the 5 in one move.
Input: board = [[1,2,3],[5,4,0]]Output: -1Explanation: No number of moves will make the board solved.
Input: board = [[4,1,2],[5,0,3]]Output: 5Explanation: 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:
boardwill be a 2 x 3 array as described above.board[i][j]will be a permutation of[0, 1, 2, 3, 4, 5].
Solution:
这道题乍一看有些蒙，但直觉上来说应该是用DFS/BFS来做，当看到要求最小步数的时候，应该要优先选择BFS。为简化操作将二维数组转化为字符串，然后用visited记录访问过的状态，用常规BFS做即可。
Code:
1 class Solution { 2 public: 3 int slidingPuzzle(vector >& board) { 4 unordered_set visited; 5 string now = ""; 6 for(int i = 0;i != 2;++i){ 7 for(int j = 0;j != 3;++j) 8 now += to_string(board[i][j]); 9 }10 if(now.compare("123450") == 0) return 0;11 visited.insert(now);12 queue > q;13 q.push(make_pair(0,now));14 vector > direction({ { 1,3},{-1,3,1},{-1,3},{-3,1},{-1,-3,1},{-3,-1}});15 while(!q.empty()){16 string front = q.front().second;17 int count = q.front().first;18 q.pop();19 int zeroindex;20 for(int i = 0;i != 6;++i){21 if(front[i] == '0')22 zeroindex = i;23 }24 for(int i = 0;i != direction[zeroindex].size();++i){25 string str = front;26 swap(str[zeroindex],str[zeroindex+direction[zeroindex][i]]);27 if(str.compare("123450") == 0) return count+1;28 if(visited.find(str) == visited.end()){29 visited.insert(str);30 q.push(make_pair(count+1,str));31 }32 }33 }34 return -1;35 }36 };