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为了解决这个问题，我们需要找到将给定网格中的“x”移动到指定位置所需的最少交换次数。我们可以使用广度优先搜索（BFS）来探索所有可能的交换状态，从而找到最短路径。

方法思路

解决代码

import java.io.BufferedReader;import java.io.IOException;import java.util.ArrayDeque;import java.util.Deque;import java.util.HashMap;import java.util.Map;class Main {    static final String START = "12345678x";    static final String[] DIRECTIONS = { "-1,0,0,-1", "1,0,0,1", "0,-1,-1,0", "0,1,1,1" };    static final int[][] D = new int[][] { {-1, 0, 0, 0}, {1, 0, 0, 0}, {0, 0, -1, 1}, {0, 0, 1, -1} };    static final String[] dx = { "-1", "1", "0", "0" };    static final String[] dy = { "0", "0", "-1", "1" };    static Map
   
     d = new HashMap<>();    static Map
    
      strsh = new HashMap<>();    static Map
     
       pre = new HashMap<>();    static Deque
      
        q = new ArrayDeque<>();    static int idx = 0;    static int res = -1;    public static void main(String[] args) throws IOException {        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));        String input = br.readLine();        input = input.replace(" ", "");        System.out.println(bfs(input));    }    private static int bfs(String start) {        q.clear();        d.clear();        strsh.clear();        pre.clear();        if (start.equals(START)) return 0;        q.add(start);        d.put(start, 0);        strsh.put(start, idx);        pre.put(start, null);        idx++;        String target = START;        while (!q.isEmpty()) {            String current = q.poll();            if (current.equals(target)) {                for (int i = res; i != -1; i = pre.get(i)) {                    res = pre.get(i);                    if (res == null) break;                }                return d.get(current);            }            int x = current.indexOf('x');            if (x == -1) continue;            int x_row = x / 3;            int x_col = x % 3;            for (int i = 0; i < 4; i++) {                int new_row = x_row + D[i][0];                int new_col = x_col + D[i][1];                if (new_row < 0 || new_row >= 3 || new_col < 0 || new_col >= 3) continue;                String[] chars = current.toCharArray();                chars[x], chars[new_row * 3 + new_col] = chars[new_row * 3 + new_col], chars[x];                String next = new String(chars);                if (!d.containsKey(next)) {                    d.put(next, d.get(current) + 1);                    strsh.put(next, idx);                    pre.put(next, current);                    idx++;                    q.add(next);                    if (next.equals(target)) {                        res = idx;                    }                }            }        }        return -1;    }}
      
     
    
   

代码解释

通过这种方法，我们可以有效地探索所有可能的交换状态，找到最少交换次数。

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