给定整数数组，在数组中找到第二大和第二小 [英] Given an array of integers, find the second largest and second smallest within the array

问题描述

我正试图提出一个函数，该函数将接受整数数组并输出第二高的数字和第二低的数字。该函数将考虑浮点数，重复项和负数。

I am trying to come up with a function that will take an array of integers and output the 2nd highest number and the 2nd smallest number. The function will take into account floats, duplicates, and negative numbers.

两个函数通过了test2之外的所有测试。

Two functions pass all the tests below except for test2.

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;
};

推荐答案

在第二个解决方案中，您的 withoutDuplicates 函数似乎是在对列表进行排序的前提下运行的（通过将元素与上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。

In your second solution, your withoutDuplicates function appears to operate on the assumption that the list is sorted (checking for duplicates by comparing an element to the previous element); however, in secondGreatLow2, you call withoutDuplicates without performing some sort of sorting.

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

If you changed the order of those two lines, solution #2 looks valid assuming you don't have any floating point mismatches, ie 3.9999999999997 != 3.99999999998

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