Question

我试图将以下c ++答案（来自上面的讨论）转换为javascript。

static bool myCompare(string a, string b){
        int i = a.find(' ');
        int j = b.find(' ');
        if(isdigit(a[i + 1]))
            if(isdigit(b[j + 1]))
                return false;       // a b are both digit logs, a == b, keep their original order
            else
                return false;       // a is digit log, b is letter log, a > b
        else
            if(isdigit(b[j + 1]))
                return true;        // a is letter log, b is digit log, a < b
            else {
                if (a.substr(i) == b.substr(j))
                    return a.substr(0,i) < b.substr(0,j); //If string part is the same, compare key
                else
                    return a.substr(i) < b.substr(j);   // a and b are both letter
            }
    }

    vector<string> reorderLogFiles(vector<string>& logs) {
        //The order of equal elements is guaranteed to be preserved in stable_sort.
        //Use sort() cannot pass the OJ. 
        stable_sort(logs.begin(), logs.end(), myCompare);
        return logs;
    }

我的解决方法如下：

var reorderLogFiles = function(logs) {
    return logs.sort(cmp);
}

var cmp = function(a, b) {
    let i = a.indexOf(' ');
    let j = b.indexOf(' ');

    if(isdigit(a[i + 1])) {
        if(isdigit(b[j + 1])) {
            // a, b digit, a == b
            return 0;
        } else {
            // a digit, b letter, b|a
            return 1;
        }
    } else {
        let condi;

        // a letter, b digit, a|b
        if(isdigit(b[j + 1])) {
            return -1;
        } else {
            // both letter
            if (a.substring(i+1) === b.substring(j+1)) {
                // start from space, all same, compare key
                condi = a.substring(0,i).localeCompare(b.substring(0,j));                
                //console.log('same', condi, a.substring(0,i), b.substring(0,j));

                return condi;
            } else {    
                condi = a.substring(i+1).localeCompare(b.substring(j+1));

                //console.log('not same', condi, a.substring(i+1), ' | ', b.substring(j+1));

                return condi;
            }
        }
    }

}

var isdigit = function(letter) {
    if(!isNaN(letter)) {
        return true;
    } else {
        return false;
    }
}

输入：

["6p tzwmh ige mc", "ns 566543603829", "ubd cujg j d yf", "ha6 1 938 376 5", "3yx 97 666 56 5", "d 84 34353 2249", "0 tllgmf qp znc", "s 1088746413789", "ys0 splqqxoflgx", "uhb rfrwt qzx r", "u lrvmdt ykmox", "ah4 4209164350", "rap 7729 8 125", "4 nivgc qo z i", "apx 814023338 8"]

我的输出：

["ubd cujg j d yf","u lrvmdt ykmox","4 nivgc qo z i","uhb rfrwt qzx r","ys0 splqqxoflgx","0 tllgmf qp znc","6p tzwmh ige mc","ns 566543603829","ha6 1 938 376 5","3yx 97 666 56 5","d 84 34353 2249","ah4 4209164350","rap 7729 8 125","apx 814023338 8","s 1088746413789"]

预期：

["ubd cujg j d yf","u lrvmdt ykmox","4 nivgc qo z i","uhb rfrwt qzx r","ys0 splqqxoflgx","0 tllgmf qp znc","6p tzwmh ige mc","ns 566543603829","ha6 1 938 376 5","3yx 97 666 56 5","d 84 34353 2249","s 1088746413789","ah4 4209164350","rap 7729 8 125","apx 814023338 8"]

我的输出和预期输出之间的差异：

"s 1088746413789"

Answer 1

您不能在JavaScript中仅通过传递 0 来实现稳定排序，而在比较其他字符串时返回的值不是{ {1}}。您将不得不将其索引存储在地图中，然后做出相应判断。我已经修改了您的代码，如下所示：

首先，我们使用其出现索引创建一个字符串映射。
后来，我们用来确定var map = {}; var reorderLogFiles = function(logs) { logs.forEach(function(value,index){ map[value] = index; }); return logs.sort(cmp); } var cmp = function(a, b) { let i = a.trim().indexOf(' '); let j = b.trim().indexOf(' '); if(isDigit(a[i+1])){ if(isDigit(b[j+1])) return map[a] - map[b]; return 1; } if(isDigit(b[j+1])){ return -1; } let cond = a.substring(i+1).localeCompare(b.substring(j+1)); if(cond == 0) return a.substring(0,i).localeCompare(b.substring(0,j)); return cond; } var isDigit = function(letter) { return !isNaN(letter); }和a何时都为数字日志。
如果两者都是字母日志，则比较忽略标识符的日志。
在发生冲突时使用标识符。

您的代码存在如下问题：

在这里，您将比较字母日志而不考虑标识符。但是，假设两者相同（在并列情况下），您无需检查其标识符的字典顺序。此外，首字母字符匹配在通常情况下也不起作用。
最佳方法是解决此问题的方法，是将字母日志和数字日志收集在不同的数组中，然后仅对字母日志进行排序，最后附加数字日志。如果许多日志是经过大量测试的数字日志，则平均情况下的性能会更好。

leetcode：渲染日志文件，无法排序

1 个答案: