2022-10-25：在一个 2 * 3 的板上（board）有 5 块砖瓦，用数字 1~5 来表示, 以及一块空缺用 0 来表示。一次 移动 定义为选择 0 与一个相邻的数字（上下左右）进行交换.

2022-10-25：在一个 2 * 3 的板上（board）有 5 块砖瓦，用数字 1~5 来表示,
以及一块空缺用 0 来表示。一次 移动 定义为选择 0 与一个相邻的数字（上下左右）进行交换.
最终当板 board 的结果是 [[1,2,3],[4,5,0]] 谜板被解开。
给出一个谜板的初始状态 board ，
返回最少可以通过多少次移动解开谜板，如果不能解开谜板，则返回 -1 。
输入：board = [[1,2,3],[4,0,5]]。
输出：1。

答案2022-10-25：

力扣773。A*算法，曼哈顿距离。
经过测试，rust的运行速度和内存占用都是最优的，go次之，java再次之。c++运行速度比java还慢了。
这道题可以用穷举打表法。

代码用rust编写。代码如下：

use std::collections::HashSet;
impl Solution {
    pub fn sliding_puzzle(m: Vec<Vec<i32>>) -> i32 {
        unsafe {
            nexts = vec![0, 0, 0];
            let mut set: HashSet<i32> = HashSet::new();
            let from =
                m[0][0] * b6 + m[0][1] * b5 + m[0][2] * b4 + m[1][0] * b3 + m[1][1] * b2 + m[1][2];
            let mut heap = Vec::new();
            // [
            // 从from出发到达当前状态，已经走了几步，
            // 从当前状态到最终状态的估计距离
            // 当前状态是什么，用数字代表
            // ]
            heap.push(vec![0, Solution::distance(from), from]);
            let mut ans = -1;
            while heap.len() > 0 {
                heap.sort_by(|a, b| (b[0] + b[1]).cmp(&(a[0] + a[1])));
                let arr = heap.pop().unwrap();
                let distance = arr[0];
                let cur = arr[2];
                if set.contains(&cur) {
                    continue;
                }
                if cur == 123450 {
                    ans = distance;
                    break;
                }
                set.insert(cur);
                let next_size = Solution::nexts(cur);
                for i in 0..next_size {
                    let next = nexts[i as usize];
                    if !set.contains(&next) {
                        heap.push(vec![distance + 1, Solution::distance(next), next]);
                    }
                }
            }
            return ans;
        }
    }

    unsafe fn nexts(from: i32) -> i32 {
        // 301245
        //  10000
        // a = 3
        let a = from / b6;
        // b = 0
        let b = (from / b5) % 10;
        // c = 1
        let c = (from / b4) % 10;
        // d = 2
        let d = (from / b3) % 10;
        // e = 4
        let e = (from / b2) % 10;
        // f = 5
        let f = from % 10;
        if a == 0 {
            nexts[0] = from + (b - a) * b6 + (a - b) * b5;
            nexts[1] = from + (d - a) * b6 + (a - d) * b3;
            return 2;
        } else if b == 0 {
            nexts[0] = from + (a - b) * b5 + (b - a) * b6;
            nexts[1] = from + (c - b) * b5 + (b - c) * b4;
            nexts[2] = from + (e - b) * b5 + (b - e) * b2;
            return 3;
        } else if c == 0 {
            nexts[0] = from + (b - c) * b4 + (c - b) * b5;
            nexts[1] = from + (f - c) * b4 + (c - f);
            return 2;
        } else if d == 0 {
            nexts[0] = from + (a - d) * b3 + (d - a) * b6;
            nexts[1] = from + (e - d) * b3 + (d - e) * b2;
            return 2;
        } else if e == 0 {
            nexts[0] = from + (b - e) * b2 + (e - b) * b5;
            nexts[1] = from + (d - e) * b2 + (e - d) * b3;
            nexts[2] = from + (f - e) * b2 + (e - f);
            return 3;
        } else {
            nexts[0] = from + (e - f) + (f - e) * b2;
            nexts[1] = from + (c - f) + (f - c) * b4;
            return 2;
        }
    }

    // 当前的数，num
    // 最终要去的数，123450
    // 返回num -> 123450 曼哈顿距离！
    fn distance(num: i32) -> i32 {
        let a = -1;
        let b = i32::abs(a);
        let mut ans = end[(num / b6) as usize][0] + end[(num / b6) as usize][1];
        ans +=
            end[((num / b5) % 10) as usize][0] + i32::abs(end[((num / b5) % 10) as usize][1] - 1);
        ans +=
            end[((num / b4) % 10) as usize][0] + i32::abs(end[((num / b4) % 10) as usize][1] - 2);
        ans +=
            i32::abs(end[((num / b3) % 10) as usize][0] - 1) + end[((num / b3) % 10) as usize][1];
        ans += i32::abs(end[((num / b2) % 10) as usize][0] - 1)
            + i32::abs(end[((num / b2) % 10) as usize][1] - 1);
        ans +=
            i32::abs(end[(num % 10) as usize][0] - 1) + i32::abs(end[(num % 10) as usize][1] - 2);
        return ans;
    }
}

const b6: i32 = 100000;

const b5: i32 = 10000;

const b4: i32 = 1000;

const b3: i32 = 100;

const b2: i32 = 10;

static mut nexts: Vec<i32> = vec![];

const end: [[i32; 2]; 6] = [[1, 2], [0, 0], [0, 1], [0, 2], [1, 0], [1, 1]];

fn main() {
    let nums = vec![vec![1, 2, 3], vec![4, 0, 5]];
    let ans = Solution::sliding_puzzle(nums);
    println!("ans = {:?}", ans);
}

struct Solution {}

执行结果如下：

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左神java代码