交换LexOrder联合查找

Question

所以我正在做面试练习问题，我遇到了这个问题:给定一个字符串字符串和指示字符串中哪些索引可以交换的配对数组，返回执行允许的交换所产生的字典顺序上最大的字符串。您可以交换索引任意次数。

示例

对于str = "abdc“和pairs = [1，4，3，4]，输出应该是swapLexOrder(str，pairs) = "dbca”。

通过交换给定的索引，您可以获得字符串："cbda“、"cbad”、"dbac“、"dbca”。此列表中按字典顺序排列的最大字符串是"dbca“。

我有一个关于寻找工会的有效答案，但我的答案太慢了：

[time limit] 4000ms (js)
0 ≤ pairs.length ≤ 5000,
pairs[i].length = 2.
1 ≤ str.length ≤ 10^4

有没有人能帮我调整一下代码来提高速度？下面是我的代码：

  function swapLexOrder(str, pairs) {
    if (!pairs.length){
        return str
    }

    answerHash = {}
    unmoved = findUnmoved(pairs, str.length)
    unionsArr = findUnions(pairs)


    for (i in unmoved){
        answerHash[unmoved[i]] = str[(unmoved[i]-1)]
    }

    unionsArr.forEach(function(union){
        letters = []
        for (i in union){
            letters.push(str[(union[i]-1)])
        }

        letters.sort()
        union.sort(function(a,b){
            return b-a
        })

        for (j in union){
            answerHash[union[j]] = letters[j]
        }
     })

    string = []
    for (keys in answerHash){
        string.push(answerHash[keys])
    }
    return string.join('')
}



//if two pairs share a number they belong in the same array
findUnions = function(pairs, unions){
   if (!unions){
       unions = [pairs[0]];
       pairs.shift();
   }else{
       if(pairs.length){
           unions.push(pairs[0])
           pairs.shift()
       }
   }

    if (!pairs.length){
        return unions
    }

    unite = true
    while (unite && pairs.length){
        unite = false
        loop1:
        for (i in unions){
            loop2:
            for (j in pairs){
                if (unions[i].includes(pairs[j][0])){
                    unions[i].push(pairs[j][1])
                    pairs.splice(j, 1)
                    unite = true
                    break loop1
                }else if (unions[i].includes(pairs[j][1])){
                    unions[i].push(pairs[j][0])
                    pairs.splice(j, 1)
                    unite = true
                    break loop1
                }
            }
        }
    }
    return findUnions(pairs, unions)
}


findUnmoved = function(pairs, length){
    range = []
    for (var i=1;i<length+1;i++){
        range.push(i);
    }
    allNum = [].concat.apply([], pairs)
    range = range.filter(function(x){
        return (!allNum.includes(x))
    })
    return range
}

这可能是我用来查找联合的函数，但我在想，也许我可以在不创建散列的情况下这样做？另外，如果你知道解决这个问题的更好的方法，我总是乐于学习新的东西。谢谢!

Answer 1

这个运行得更快。

function swapLexOrder(str, pairs) {
//Turn pairs into edge lists: O(n+m)
var graph = new Array(str.length).fill(0).map(e=>[]);
for(var pair of pairs) {
    graph[pair[0]-1].push(pair[1]-1);
    graph[pair[1]-1].push(pair[0]-1);
}

//Build all the ccs with dfs: O(n+m)
var ccs = [], ccnum = 0;
for(var c in str) {
    if(ccs[c])
        continue;
    ccs[c] = ++ccnum;
    var dfs = [...graph[c]];
    while(dfs.length) {
        var d = dfs.shift();
        if(ccs[d])
            continue;
        ccs[d] = ccnum;
        dfs.push(...graph[d]);
    }
}

//Group words by ccs: O(n)
var ccWords = new Array(ccnum).fill(0).map(e=>[]);
for(var c in str) {
    ccWords[ccs[c]-1].push(str[c]);
}

//Sort all words: O(n log n)
ccWords.map(e=>e.sort());

//Build the new string: O(n)
var output = "";
for(var c in str) {output += ccWords[ccs[c]-1].pop(); }
    return output;
}