javascript - 给定一个整数数组，找到数组中第二大和第二小的整数

Question

我正在尝试想出一个函数，它将接受一个整数数组并输出第二大数字和第二小数字。该函数将考虑 float 、重复数和负数。

两个函数通过了除 test2 之外的所有测试。

var test1 = [7, 7, 12, 98, 106]
  answer1 = {2nd Min: 12, 2nd Max: 98}

var test2 = [5, 23, -112, 6, 70, 70, -112]
  answer2 = {2nd Min: 5, 2nd Max: 23}

var test3 = [-22, 22]
  answer3 = {2nd Min: 22, 2nd Max: -22}

var test4 = [10, 89, 3]
   answer = {2nd Min: 10, 2nd Max: 10}

var test5 = [10.4, -12.09, .75, 22]
  answer3 = {2nd Min: 0.75, 2nd Max: 10.4}

 /*/          \ \
 ---SOLUTION 1---
 \ \          /*/
function secondGreatLow1(arr) {
  //make copy of array because it will be spliced in the following functions
  var arrCopy = arr.slice();
  //push returned values of each function into this the answer array
  var answer = []
  answer.push(secondMin(arrCopy));
  answer.push(secondMax(arrCopy));
  return answer;
};

//helper function 1
var secondMin = function (arr){
  var arrCopy = arr.slice();
  //check length of array
  if (arr.length == 2) {
    return arr[1];
  } else {
    var min = Math.min.apply(null, arrCopy);
    arrCopy.splice(arrCopy.indexOf(min), 1);
    //check for duplicates
    for (var i = 0; i < arrCopy.length; i++) {
      if (arrCopy.indexOf(min) === -1) {
        //.apply is used for arrays
        return Math.min.apply(null, arrCopy);
      } else {
        arrCopy.splice(arrCopy.indexOf(min), 1);
        return Math.min.apply(null, arrCopy);
      }
    };
  }
};

//helper function 2
var secondMax = function (arr){
  var arrCopy = arr.slice();
  if (arr.length == 2) {
    return arr[0];
  } else {
    var max = Math.max.apply(null, arrCopy);
    arrCopy.splice(arrCopy.indexOf(max), 1);
    //check for duplicates
    for (var i = 0; i < arrCopy.length; i++) {
      if (arrCopy.indexOf(max) === -1) {
        return Math.max.apply(null, arrCopy);
      } else {
        arrCopy.splice(arrCopy.indexOf(max), 1);
        return Math.min.apply(null, arrCopy);
      }
    };
  }
};



 /*/          \ \
 ---SOLUTION 2---
 \ \         /*/
function secondGreatLow2 (numbers) {
  var arr = withoutDuplicates(numbers);
  arr.sort(function(a,b) { return a-b; });
  return arr[1] + ' ' + arr[arr.length-2];
};

// helpers
var withoutDuplicates = function(arr) {
  var out = [];
  for(var i=0; i<arr.length; i++) {
    if(i === 0 || arr[i] !== arr[i-1]) {
      out.push(arr[i]);
    }
  }
  return out;
};

Answer 1

在您的第二个解决方案中，您的 withoutDuplicates 函数似乎在列表已排序的假设下运行(通过将一个元素与前一个元素进行比较来检查重复项)；但是，在 secondGreatLow2 中，您调用 withoutDuplicates 而不执行某种排序。

如果您更改了这两行的顺序，则解决方案 #2 看起来有效假设您没有任何浮点不匹配，即 3.9999999999997 != 3.99999999998