Question

我有一个数组，它包含教程中的每一步，并且它对应的git commit sha。我想选择每章的最后一步，对于以下示例，它对应于1.1，2.5和3.4。我目前正在使用Typescript中的 let ... of 迭代器来分析步骤。

[1]     step: '3.4' },
[1]   Commit {
[1]     hash: '52f7762e2a2d901b9d849e70679237fad8f0c05f',
[1]     step: '3.3' },
[1]   Commit {
[1]     hash: '6ef63c21cd7f42fe5f190fb7bd5a7411528b944e',
[1]     step: '3.2' },
[1]   Commit {
[1]     hash: 'fe7bd36851672a24c0cd205f0763d52e6abfa6e1',
[1]     step: '3.1' },
[1]   Commit {
[1]     hash: '13b948cc246b3c9b383c4be24ca0ba0a7c072e67',
[1]     step: '2.5' },
[1]   Commit {
[1]     hash: '679bc61d53a59ad10c0398f2faecd7a4052689dc',
[1]     step: '2.4' },
[1]   Commit {
[1]     hash: '76f57e963afb885fed48b1646fd50025269061ac',
[1]     step: '2.3' },
[1]   Commit {
[1]     hash: 'fd368fed4f47b9686e855b2a76e53dae5880ef69',
[1]     step: '2.2' },
[1]   Commit {
[1]     hash: 'a70e6f556640db53f1ef3acba28c42f582d45890',
[1]     step: '2.1' },
[1]   Commit {
[1]     hash: '95854756de842ff45ebbf9d3703cc7eff1557d5a',
[1]     step: '1.1' },

这样做的方法是什么？

我目前的方法需要循环遍历所有提交以查找有多少章节，然后使用step = chapter.x过滤掉所有结果。然后找到该数组中最大的x。然后重复所有章节。对于我想要实现的目标，这似乎非常笨拙。

Answer 1

我会看一下像下划线max()这样的东西，它允许你传入数组，一个测试最大值的函数，并返回最大值。

如果这是一个项目，或者由于某种原因你不能使用库，那么一个简单的for循环就足够了。

Answer 2

这是线性的最坏情况复杂，所以我能想到的最简单的方法（在简单的javascript中）是：

＆＃13;

function maxChapters(commits) {
  var result = [];
  var recent = NaN;
  for (var i = 0; i < commits.length; i++) {
    var chapter = Math.floor(commits[i]['step']);
    if (chapter != recent) {
      result.push(commits[i]['step']);
      recent = chapter;
    }
  }
  return result;
}

var commits = [{
  hash: '52f7762e2a2d901b9d849e70679237fad8f0c05f',
  step: '3.3' 
}, {
  hash: '6ef63c21cd7f42fe5f190fb7bd5a7411528b944e',
  step: '3.2' 
}, {
  hash: 'fe7bd36851672a24c0cd205f0763d52e6abfa6e1',
  step: '3.1' 
}, {
  hash: '13b948cc246b3c9b383c4be24ca0ba0a7c072e67',
  step: '2.5' 
}, {
  hash: '679bc61d53a59ad10c0398f2faecd7a4052689dc',
  step: '2.4' 
}, {
  hash: '76f57e963afb885fed48b1646fd50025269061ac',
  step: '2.3' 
},{
  hash: 'fd368fed4f47b9686e855b2a76e53dae5880ef69',
  step: '2.2' 
}, {
  hash: 'a70e6f556640db53f1ef3acba28c42f582d45890',
  step: '2.1' 
}, {
  hash: '95854756de842ff45ebbf9d3703cc7eff1557d5a',
  step: '1.1' 
}]

console.log(maxChapters(commits));

＆＃13;

Answer 3

有很多方法可以解决这个问题。我使用节号来抵消数组中的索引。这节省了对数组的迭代（每章一次迭代而不是迭代每一步），以换取必须对每个元素进行更多操作（拆分步骤字符串并将步骤转换为int）。你的旅费可能会改变。与其他答案一样，这是一个简单的旧JavaScript解决方案。

＆＃13;

//Only this function is the actual solution
function findLastSections(list) {
  var chapterEnds = [];
  var currCommit;
  var currIndex = 0;
  
  //Exit clause is when my index goes out-of-bounds.
  while(currIndex < list.length) {
    //I took the goal of trying to only iterate once per chapter.
    console.log('processing element: ' + currIndex);
    //get the last commit for the chapter
    currCommit = list[currIndex];
    //add its step to our return array.
    chapterEnds.unshift(currCommit.step);
    //split the step on the decimal, and parseInt so that I can do arithmetic on it. Add it to the index. This effectively shifts my position in the array by the length of the current chapter.
    currIndex += parseInt(currCommit.step.split('.')[1]);
  }
  console.log('returning: '); 
  console.log(chapterEnds);
  return chapterEnds;
}

//Everything below this line is just setup stuff to approximate your environment.

//Call to the function.
findLastSections(getCommitList());

 // Constructor function in place of Typescript class
function Commit(hash, step) {
  this.hash = hash;
  this.step = step;
}

function getCommitList() {
  //just did this to get it hoisted, so it's not crammed into the top of the snippet.
  return [
      new Commit (
     '52f7362e2a2d901b9d849e70679237fad8f0c05f',
     '3.4'),
      new Commit (
     '52f7762e2a2d901b9d849e70679237fad8f0c05f',
     '3.3'),
   new Commit (
     '6ef63c21cd7f42fe5f190fb7bd5a7411528b944e',
     '3.2'),
   new Commit (
     'fe7bd36851672a24c0cd205f0763d52e6abfa6e1',
     '3.1'),
   new Commit (
     '13b948cc246b3c9b383c4be24ca0ba0a7c072e67',
     '2.5'),
   new Commit (
     '679bc61d53a59ad10c0398f2faecd7a4052689dc',
     '2.4'),
   new Commit (
     '76f57e963afb885fed48b1646fd50025269061ac',
     '2.3'),
   new Commit (
     'fd368fed4f47b9686e855b2a76e53dae5880ef69',
     '2.2'),
   new Commit (
     'a70e6f556640db53f1ef3acba28c42f582d45890',
     '2.1'),
   new Commit (
     '95854756de842ff45ebbf9d3703cc7eff1557d5a',
     '1.1')
];
}

＆＃13;

查找数组中每个整数的最大小数

3 个答案: