javascript - 我的 javascript 代码返回最常见的字母有什么问题？

Question

我正在尝试编写一个解决方案，返回字符串中最常见的字母，以及该字母出现的次数。 我想出的解决方案是循环遍历字符串，一次将一个字母保存在一个名为 letter 的变量中，并将所有其他字母与此进行比较。然后一旦字符串结束，循环将保持第二个字母并重复。

我遇到的问题是它每次都只返回字符串中的第一个字母。它得到正确的匹配数，但忽略了其他字母有更多匹配的事实。

我做错了什么？

var match = 0;
var matchCount = 0;
var letter = 0;
var count = 0;
indx1 = 0;
indx2 = 0;

function problemTen(a) {
  while (indx1 < a.length) {
    letter = a[indx1]; //hold the first letter in the string in the letter variable

    while (indx2 < a.length) { //now rotate through every other letter to compare
      if (a[indx2] == letter) { //compare 
        count += 1; // if there is a match, add one to the count variable to hold the number of matches
      }
      indx2 += 1; //cycle through the rest of the string  
    }
    if (matchCount === 0 || count > matchCount) { // if it’s the first time around,     or if this letter had more matches than the previous letter
      match = letter; // hold the letter in the match variable
      matchCount = count; //hold the number of matches in the count     variable
    }
    indx1 += 1; //cycle through the first variable that you compare
  }
  return [match, matchCount]; //return results
}

console.log(problemTen("astrings"));

编辑:这是我想出的解决方案。

function problemTen (a) {
  var match = 0;
  var matchCount = 0;
  var letter;

  for(var indx1 = 0; indx1<a.length; indx1++) {
    letter = a[indx1];              
    var count = 1;                  

    for(var indx2 = indx1 + 1; indx2<a.length; indx2++) {         
      if(a[indx2] == letter) {        
        count +=1;                 
      }
    }
    if(matchCount === 0 || count>matchCount) { 
      match = letter;                  
      matchCount =count;               
    }   
  }
  return [match, matchCount];       
}

console.log(problemTen("iamatinystring"));

Answer 1

也许检查这个解决方案，它的时间复杂度与你试图实现的那个相比是线性的，它是四倍的:

function problemTen(a) {
  var array = a.split('');
  var max = 0;
  var letter;

  var counter = array.reduce(function(memo, item) {
    memo[item] = memo[item] ? memo[item] + 1 : 1;
    if (memo[item] > max) {
      max = memo[item];
      letter = item;
    }
    return memo;
  }, {});

  return [letter, max];
}

console.log(problemTen("astrings"));

如果你想坚持你的解决方案，问题是，当你再次运行外部 while 循环时，你并没有重置 indx2 和 count 变量，因此，indx2 = a.length 已经存在，而内部 while循环不再运行来计算字母。添加这些重置后，您的代码应该是这样的:

function problemTen(a) {
  var match = 0;
  var matchCount = 0
  var letter = 0;
  var count = 0;
  indx1 = 0;
  indx2 = 0;

  while (indx1 < a.length) {
    letter = a[indx1]; //hold the first letter in the string in the letter variable

    while (indx2 < a.length) { //now rotate through every other letter to compare
      if (a[indx2] === letter) { //compare 
        count += 1; // if there is a match, add one to the count variable to hold the number of matches
      }
      indx2 += 1; //cycle through the rest of the string  
    }
    if (matchCount === 0 || count > matchCount) { // if it’s the first time around,     or if this letter had more matches than the previous letter
      match = letter; // hold the letter in the match variable
      matchCount = count; //hold the number of matches in the count     variable
    }
    indx1 += 1; //cycle through the first variable that you compare

    // HERE WE RESET indx2 and match to 0
    indx2 = 0;
    match = 0;

  }
  return [match, matchCount]; //return results
}

console.log(problemTen("astrings"));

其实要达到O(n)的时间复杂度，不一定要用reduce，一个循环也可以，而且肯定会快很多，我只是更喜欢函数式编程，因为它的结果更好的可读性:

function problemTen(a) {
  var max = 0;
  var letter;
  var counter = {};

  for (var i = 0, l = a.length; i<l; i++) {
    counter[a[i]] = counter[a[i]] ? counter[a[i]] + 1 : 1;
    if (counter[a[i]] > max) {
      max = counter[a[i]];
      letter = a[i];
    }
  }

  return [letter, max];
}

console.log(problemTen("astrings"));

希望这对您有所帮助。