题目大意:给你一个2×3的棋盘,放着0-5。每一步0可以和上下左右的一个数交换。问需要多少步可以构成123450的棋盘状态。
Problem:
On a 2×3 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:
1 2 3 |
Input: board = [[1,2,3],[4,0,5]] Output: 1 Explanation: Swap the 0 and the 5 in one move. |
1 2 3 |
Input: board = [[1,2,3],[5,4,0]] Output: -1 Explanation: No number of moves will make the board solved. |
1 2 3 4 5 6 7 8 9 10 |
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]] |
1 2 |
Input: board = [[3,2,4],[1,5,0]] Output: 14 |
Solution: BFS
Time complexity: O(6!)
Space complexity: O(6!)
C++
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// Author: Huahua // Running time: 9 ms class Solution { public: int slidingPuzzle(vector<vector<int>>& board) { constexpr int kRows = 2; constexpr int kCols = 3; string goal; string start; for (int i = 0; i < board.size(); ++i) for (int j = 0; j < board[0].size(); ++j) { start += (board[i][j] + '0'); goal += (i * kCols + j + 1) % (kRows * kCols) + '0'; // 12345...0 } if (start == goal) return 0; constexpr int dirs[4][2] = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}}; set<string> visited{start}; int steps = 0; queue<string> q; q.push(start); while (!q.empty()) { ++steps; int size = q.size(); while (size-- > 0) { string s = q.front(); q.pop(); int p = s.find('0'); int y = p / kCols; int x = p % kCols; for (int i = 0; i < 4; ++i) { int tx = x + dirs[i][0]; int ty = y + dirs[i][1]; if (tx < 0 || ty < 0 || tx >= kCols || ty >= kRows) continue; int pp = ty * kCols + tx; string t(s); swap(t[p], t[pp]); if (visited.count(t)) continue; if (t == goal) return steps; visited.insert(t); q.push(t); } } } return -1; } }; |
Simplified, only works on 3×2 board
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// Author: Huahua // Running time: 9 ms class Solution { public: int slidingPuzzle(vector<vector<int>>& board) { const string goal = "123450"; string start; for (int i = 0; i < board.size(); ++i) for (int j = 0; j < board[0].size(); ++j) start += (board[i][j] + '0'); if (start == goal) return 0; const vector<vector<int>> idx{{1, 3}, {0, 2, 4}, {1, 5}, {0, 4}, {1, 3, 5}, {2, 4}}; set<string> visited{start}; int steps = 0; queue<pair<string, int>> q; q.emplace(start, start.find('0')); while (!q.empty()) { ++steps; int size = q.size(); while (size-- > 0) { const auto& p = q.front(); q.pop(); for (int index : idx[p.second]) { string t(p.first); swap(t[p.second], t[index]); if (visited.count(t)) continue; if (t == goal) return steps; visited.insert(t); q.emplace(t, index); } } } return -1; } }; |