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-rw-r--r--sliding_puzzle.cpp53
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diff --git a/sliding_puzzle.cpp b/sliding_puzzle.cpp
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-// 773. Sliding Puzzle
-#include "leetcode.h"
-
-/*
- * on an empty 2 x 3 board there are five tiles labeled from 1 to 5 and an empty
- * square represented by 0. a move consists of choosing 0 and a 4-directionall
- * 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 the 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
- */
-
-class Solution {
-public:
- int slidingPuzzle(vector<vector<int>> &board) {
- string begin = "", end = "123450";
- vector<vector<int>> moves = {{1, 3}, {0, 2, 4}, {1, 5},
- {0, 4}, {1, 3, 5}, {2, 4}};
- for (int i = 0; i < board.size(); i++)
- for (int j = 0; j < board[0].size(); j++)
- begin += to_string(board[i][j]);
- unordered_set<string> vis;
- vis.insert(begin);
- queue<pair<string, int>> qp;
- qp.push({begin, 0});
- while (!qp.empty()) {
- auto p = qp.front();
- qp.pop();
- string curr = p.first;
- int dist = p.second, idx = curr.find("0");
- if (curr == end)
- return dist;
- for (auto &i : moves[idx]) {
- string next = curr;
- swap(next[idx], next[i]);
- if (!vis.count(next)) {
- vis.insert(next);
- qp.push({next, dist + 1});
- }
- }
- }
- return -1;
- }
-};
-
-int main() {
- Solution obj;
- vvd(int) b1 = {{1, 2, 3}, {4, 0, 5}}, b2 = {{1, 2, 3}, {5, 4, 0}},
- b3 = {{4, 1, 2}, {5, 0, 3}};
- printf("%d\n", obj.slidingPuzzle(b1)); // expect 1
- printf("%d\n", obj.slidingPuzzle(b2)); // expect -1
- printf("%d\n", obj.slidingPuzzle(b3)); // expect 5
-}