这就是我的解。它已经在leetcode上被接受。我在今天的面试中遇到了问题。但我当时没有解决它。 我又想了想。加0使两个数组的长度相等。然后比较。

var compareVersion = function(version1, version2) { let arr1 = version1.split('.').map(Number); let arr2 = version2.split('.').map(Number); let diff = 0; if (arr1.length > arr2.length){ diff = arr1.length - arr2.length; while (diff > 0){ arr2.push(0); diff--; } } else if (arr1.length < arr2.length){ diff = arr2.length - arr1.length; while (diff > 0){ arr1.push(0); diff--; } } let i = 0; while (i < arr1.length){ if (arr1[i] > arr2[i]){ return 1; } else if (arr1[i] < arr2[i]){ return -1; } i++; } return 0; };

你可以使用带有选项的String#localeCompare

sensitivity Which differences in the strings should lead to non-zero result values. Possible values are: "base": Only strings that differ in base letters compare as unequal. Examples: a ≠ b, a = á, a = A. "accent": Only strings that differ in base letters or accents and other diacritic marks compare as unequal. Examples: a ≠ b, a ≠ á, a = A. "case": Only strings that differ in base letters or case compare as unequal. Examples: a ≠ b, a = á, a ≠ A. "variant": Strings that differ in base letters, accents and other diacritic marks, or case compare as unequal. Other differences may also be taken into consideration. Examples: a ≠ b, a ≠ á, a ≠ A. The default is "variant" for usage "sort"; it's locale dependent for usage "search". numeric Whether numeric collation should be used, such that "1" < "2" < "10". Possible values are true and false; the default is false. This option can be set through an options property or through a Unicode extension key; if both are provided, the options property takes precedence. Implementations are not required to support this property.

var版本=[" 2.0.1”、“2.0”、“1.0”、“1.0.1”,“2.0.0.1”); 版本。sort((a, b) => a.localeCompare(b, undefined， {numeric: true，灵敏度:'base'})); console.log(版本);

我们的想法是比较两个版本，并知道哪个是最大的。我们删除“。”，并将向量的每个位置与其他位置进行比较。

// Return 1  if a > b
// Return -1 if a < b
// Return 0  if a == b

function compareVersions(a_components, b_components) {

   if (a_components === b_components) {
       return 0;
   }

   var partsNumberA = a_components.split(".");
   var partsNumberB = b_components.split(".");

   for (var i = 0; i < partsNumberA.length; i++) {

      var valueA = parseInt(partsNumberA[i]);
      var valueB = parseInt(partsNumberB[i]);

      // A bigger than B
      if (valueA > valueB || isNaN(valueB)) {
         return 1;
      }

      // B bigger than A
      if (valueA < valueB) {
         return -1;
      }
   }
}

以下是我的解决方案，适用于任何深度的任何版本。

自动处理数字+点问题。如果不是这样，函数存在，控制台日志将给出undefined而不是true, false或true。

自动处理尾随零问题。

任何可能的地方都存在自动继电器。

自动向后兼容旧浏览器。

function checkVersion (vv,vvv){ if(!(/^[0-9.]*$/.test(vv) && /^[0-9.]*$/.test(vvv))) return; va = vv.toString().split('.'); vb = vvv.toString().split('.'); length = Math.max(va.length, vb.length); for (i = 0; i < length; i++) { if ((va[i]|| 0) < (vb[i]|| 0) ) {return false; } } return true;} console.log(checkVersion('20.0.0.1' , '20.0.0.2')); console.log(checkVersion(20.0 , '20.0.0.2')); console.log(checkVersion('20.0.0.0.0' , 20)); console.log(checkVersion('20.0.0.0.1' , 20)); console.log(checkVersion('20.0.0-0.1' , 20));

看看这篇博客文章。此函数适用于数字版本号。

function compVersions(strV1, strV2) {
  var nRes = 0
    , parts1 = strV1.split('.')
    , parts2 = strV2.split('.')
    , nLen = Math.max(parts1.length, parts2.length);

  for (var i = 0; i < nLen; i++) {
    var nP1 = (i < parts1.length) ? parseInt(parts1[i], 10) : 0
      , nP2 = (i < parts2.length) ? parseInt(parts2[i], 10) : 0;

    if (isNaN(nP1)) { nP1 = 0; }
    if (isNaN(nP2)) { nP2 = 0; }

    if (nP1 != nP2) {
      nRes = (nP1 > nP2) ? 1 : -1;
      break;
    }
  }

  return nRes;
};

compVersions('10', '10.0'); // 0
compVersions('10.1', '10.01.0'); // 0
compVersions('10.0.1', '10.0'); // 1
compVersions('10.0.1', '10.1'); // -1

replace()函数只替换字符串中的第一个出现项。我们来替换。与,。然后全部删除。然后做，to。再次将其解析为float。

for(i=0; i<versions.length; i++) {
    v = versions[i].replace('.', ',');
    v = v.replace(/\./g, '');
    versions[i] = parseFloat(v.replace(',', '.'));
}

最后，排序:

versions.sort();