javascript - JavaScript 中的 "Look and say sequence"

Question

1
11
12
1121
122111
112213
122211
....
I was trying to solve this problem. It goes like this. I need to check the former line and write: the number and how many time it was repeated.
ex. 1 -> 1(number)1(time)

var antsArr = [[1]];
var n = 10;

for (var row = 1; row < n; row++) { 
  var lastCheckedNumber = 0;
  var count = 1;
  antsArr[row] = [];

  for (var col = 0; col < antsArr[row-1].length; col++) {

    if (lastCheckedNumber == 0) {   
        lastCheckedNumber = 1;
        antsArr[row].push(lastCheckedNumber);

    } else {

        if (antsArr[row-1][col] == lastCheckedNumber) {
            count++;

        } else {
            lastCheckedNumber = antsArr[row-1][col];
        }

    }

  }
  antsArr[row].push(count);
  antsArr[row].push(lastCheckedNumber);
}



for (var i = 0; i < antsArr.length; i++) {
  console.log(antsArr[i]);
}

我从 2 天前就开始关注这个了。自己很难解决。我知道这对你们来说真的是基本代码。但是，如果有一个真正热心的人帮助我，我会很高兴! :>

Answer 1

试试这个:

JSFiddle Sample

function lookAndSay(seq){
    var prev = seq[0];
    var freq = 0;
    var output = [];
    seq.forEach(function(s){
        if (s==prev){
            freq++;
        }
        else{
            output.push(prev);
            output.push(freq);
            prev = s;
            freq = 1;
        }
    });
    output.push(prev);
    output.push(freq);
    console.log(output);
    return output;
}

// Sample: try on the first 11 sequences
var seq = [1];
for (var n=0; n<11; n++){
    seq = lookAndSay(seq);
}

快速解释

输入序列是一个包含序列中所有数字的简单数组。函数遍历序列中的元素，统计当前出现的次数。当它遇到一个新数字时，它会将先前出现的数字连同频率一起推送到输出。

保持迭代直到结束，确保将最后出现的数字和频率添加到输出中，仅此而已。