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原文链接：blog.csdn.net/roufoo/arti…
Sliding Puzzle II 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 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 Challenge How to optimize the memory? Can you solve it with A* algorithm?
解法1：BFS 注意：这种棋盘类的题目用BFS都要把状态转换为一维来方便处理。 代码如下：
class Solution {
public:
/**
* @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
*/
int minMoveStep(vector<vector<int>> &init_state, vector<vector<int>> &final_state) {
string beginStr = getStr(init_state);
string endStr = getStr(final_state);
if (beginStr == endStr) return 0;
unordered_set<string> visited;
queue<pair<string, int>> q; //string, pos
//pair<string, int> node = make_pair(beginStr, getZeroPos(init_state));
//pair<string, int> node(beginStr, getZeroPos(init_state));
pair<string, int> node = {beginStr, getZeroPos(init_state)};
q.push(node);
vector<int> dx = {1, -1, 0, 0};
vector<int> dy = {0, 0, 1, -1};
int step = 1;
while(!q.empty()) {
int qSize = q.size();
for (int i = 0; i < qSize; ++i) {
node = q.front();
string frontStr = node.first;
q.pop();
int x = node.second / 3;
int y = node.second % 3;
for (int j = 0; j < 4; ++j) {
int newX = x + dx[j];
int newY = y + dy[j];
if (newX >= 0 && newX < 3 && newY >= 0 && newY < 3) {
int newPos = newX * 3 + newY;
swap(frontStr[node.second], frontStr[newPos]);
if (visited.find(frontStr) == visited.end()) {
if (frontStr == endStr) {
return step;
} else {
visited.insert(frontStr);
q.push({frontStr, newPos});
}
}
frontStr = node.first;
}
}
}
step++;
}
return -1;
}
private:
string getStr(vector<vector<int>> &vv) {
string str = "";
for (int i = 0; i < 3; ++i) {
for (int j = 0; j < 3; ++j) {
str.push_back('0' + vv[i][j]);
}
}
return str;
}
int getZeroPos(vector<vector<int>> &vv) {
int pos = 0;
for (int i = 0; i < 3; ++i) {
for (int j = 0; j < 3; ++j) {
if (vv[i][j] == 0) {
return i * 3 + j;
}
}
}
return -1;
}
};
解法2：A*算法。 TBD。
