思路：广度搜索+去除重复步骤

//存放走过了的方案
let dic = new Set();
const target = '12345678x';

let left = x => x % 3 ? x - 1 : null;
let top = x => x >= 3 ? x - 3 : null;
let right = x => x % 3 < 2 ? x + 1 : null;
let bottom = x => x < 6 ? x + 3 : null;
let move = [left, top, right, bottom];

let swap = (arr, a, b) => {
    let t = arr[a];
    arr[a] = arr[b], arr[b] = t;
}

let bfs = arr => {
    let queue = [[arr, arr.indexOf('x')]];//存放当前网格状态与x位置
    let step = -1;
    while (queue.length > 0) {
        step++;
        let next = [];
        for (let i = 0; i < queue.length; i++) {
            let act = queue[i][0].join('');
            let x = queue[i][1];
            if (dic.has(act)) continue; //去重
            if (act === target) return step;
            dic.add(act);
            for (let j = 0; j < 4; j++) { //4个方向
                let y = move[j](x);
                if (y !== null) {
                    let copy = [].concat(queue[i][0]);
                    swap(copy, x, y);
                    next.push([copy, y]);
                }
            }
        }
        queue = next;
    }
    return -1;
}

let buf = '';
process.stdin.on('readable', function () {
    let chunk = process.stdin.read();
    if (chunk) buf += chunk.toString();
});
let getInputNums = line => line.split(' ').filter(s => s !== '').map(x => parseInt(x));
let getInputStr = line => line.split(' ').filter(s => s !== '');
process.stdin.on('end', function () {
    buf.split('\n').forEach(function (line, lineIdx) {
        if (lineIdx === 0) {
            let a = getInputStr(line);
            console.log(bfs(a));
        }
    });
});