javascript – 在元素数组中查找重复的元素系列

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我有这样的数组

 var randomArray = [1,2,1,1,1,1,0,2,1,2,3,10,12,54,10,12] etc..

我可以删除重复的元素或在此查找重复的元素.但我想记录在数组中重复的所有重复元素序列.这是我尝试过的代码,但它运行到无限循环中

  for (i = 0; i < randomLength; i++) {
    var cycle = [i],
    flag = 0,
    start = i;
    for (var j = i + 1; j < randomLength; j++) {
       if (randomArray[i] == randomArray[j]) {
         cycle.push(randomArray[j]);
         while (i <= j) {
            if (randomArray[i + 1] == randomArray[j + 1]) {
                cycle.push(randomArray[j + 1]);
            }
            i = i + 1;
            j = j + 1;
         }
         console.log(cycle);
       }
       i = start;
    }
   i = start;
 }  

它应该归还我.而且我不想正则表达式做同样的事情

1,2
1,1
10,12

If array is ["a","d","z","e","g","h","a","d","z"]  

然后

output would be "a","d","z"

它应该是最佳解决方案.请建议我这个.至少更正我当前的代码..

解决方法:

Here is my solution,就像@robert国王一样(正如我自己解决问题后发现的那样),除了我的完成(已经能够不计算重叠模式)和优化(尽我所能).

此外,返回一个对象的地图,因此你可以枚举它,只能拉出大小为X的图形,或重复Y次的图形等.

以下行(具有以下功能)

getPatterns([1,2,1,1,1,1,0,2,1,2,3,10,12,54,10,12]).showRepeated();

会导致这个;

1 2 found 2 times
2 1 found 2 times
1 1 found 2 times
10 12   found 2 times

码

function getPatterns(input, generateAll) {
    var patternMap = new getPatterns.presentation();

    var generated = [];
    var patternObj;
    var start;
    //for each item
    for (var index = 0; index < input.length; ++index) {
        //open a new slot for a new pattern start at this index
        generated.push('');

        start = 0;
        //unless told to generate all
        //skip patterns that cant possibly be repeated
        //(i.e. longer than half the input length)
        if (!generateAll && generated.length > input.length / 2)
            start = generated.length - Math.floor(input.length / 2);

        //test patterns we have generated for this index
        for (var index2 = start; index2 < generated.length; ++index2) {
            //generate a fresh lot of patterns for this index
            generated[index2] += ' ' + input[index];

            //unless told to generate all, dismiss patterns of length 1
            if (!generateAll && index2 == generated.length - 1)
                break;

            //try to fetch a pre-existing pattern, O(1)
            patternObj = patternMap[generated[index2]];
            //if this is a new pattern
            if (!patternObj) {
                //generate an object
                patternMap[generated[index2]] = {
                    lastSeen : index,
                    count : 1,
                    size : generated.length - index2
                };
                continue;
            }

            //unless told to generate all, skip patterns that overlap with themselves
            if (!generateAll && index - patternObj.lastSeen < patternObj.size)
                continue;

            //this pattern has repeated! update the object data
            ++patternObj.count;
            patternObj.lastSeen = index;
        }
    }

    return patternMap;
}
//just for a function prototype
getPatterns.presentation = function() {};
getPatterns.presentation.prototype = {
    showRepeated : function() {
        var patternObj;
        for (var pattern in this) {
            patternObj = this[pattern];
            if (patternObj.count > 1)
                console.log(pattern + '\tfound ' + patternObj.count + ' times');
        }
    }
};

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