在给定的一组数字中找到时间最少的第三小数字

Question

Here's a working algorithm that finds the third smallest number in a given set of numbers.

I was looking for another solutions to the given requirement with less time complexity.

Here's the working code:

 Numbers = [3,2,55,-10,-55,5,3,2,1,-5,33,9,-1,4,5]; var x = 0; var y = 0; function FindThirdSmallestNumber() { for(var i=0;i<Numbers.length;i++) { if (Numbers[i] > Numbers[i+1]) { x = Numbers[i]; y = Numbers[i+1]; Numbers[i] = y; Numbers[i+1] = x; i=-1; } else { // } } console.log(Numbers[2]); } FindThirdSmallestNumber(); 

Answer 1

This one should be a lot simpler. Also not sure about this being any faster but in the most simple/obvious cases less code = better performance.

I just sort the array ascending and get the value based on index. So with this code you can get any place; lowest, second lowest, third lowest, etc as long as your index does not go out of range.

 const input = [3,2,55,-10,-55,5,3,2,1,-5,33,9,-1,4,5]; function getLowestByRank(data, rank) { data.sort(function(a, b){ return a - b }); return data[rank - 1]; } console.log(getLowestByRank(input, 3)) console.log(getLowestByRank(input, 2)) console.log(getLowestByRank(input, 4)) 

Answer 2

Not sure if this is any faster but it's shorter:

//Use a custom sort function and pass it to the .sort() method
Numbers = Numbers.sort(function(x, y){ return x - y; });
if(Numbers.length > 2){
    //At this point, the 3rd item in the array should be the 3rd lowest
    console.log(Numbers[2]);
}else {
    console.log("There are not 3 numbers in the array.");
}

Answer 3

One option would be to have a separate sorted array of the three smallest numbers so far. Whenever you run across a number smaller than the 3rd smallest (the last in the sorted array), reassign that 3rd index to the new number, and sort it again:

 const numbers = [3, 2, 55, -10, -55, 5, 3, 2, 1, -5, 33, 9, -1, 4, 5]; const sortNumeric = (a, b) => a - b; function FindThirdSmallestNumber(input) { const [smallestSoFar, numbers] = [input.slice(0, 3), input.slice(3)]; smallestSoFar.sort(sortNumeric); numbers.forEach((num) => { if (num < smallestSoFar[2]) { smallestSoFar[2] = num; smallestSoFar.sort(sortNumeric); } }); return smallestSoFar[2]; } console.log(FindThirdSmallestNumber(numbers)); 

Note that implementations that sort the whole array (as other answers do) is O(N log N) , while sort ing here is only ever done on an array of size 3, which is significantly less complex ( O(N log 3) I think, which is equivalent to O(N) )

Answer 4

You can use Math.min function to determine the minimum until you find your X smallest number:

Numbers = [3,2,55,-10,-55,5,3,2,1,-5,33,9,-1,4,5];

function FindSmallestNumber(arr, limit) {
   var min = '';
   for(var counter = 1; counter <= limit; counter ++) {
      min = Math.min.apply(Math, arr);
      arr.splice(arr.indexOf(min), 1);
   }   
   console.log(min);
}

FindSmallestNumber(Numbers, 3); //3rd smallest number

Answer 5

I think sorting the array is likely the fastest method, but perhaps you want avoid the built–in sort. An alternative is as Chris Li suggests: iterate over the values and just keep the 3 lowest, then return the highest of the three.

I assumed you want to avoid built-in methods, so this only uses some basic array methods and does everything else manually.

 // Avoid Math.max function getMax(arr) { var max = -Infinity; for (var i=0, iLen=arr.length; i<iLen; i++) { if (arr[i] > max) max = arr[i]; } return max; } // If data.length < 4, just returns largest value // Iterates over data once function get3rdSmallest(data) { // Helper to insert value in order // Expects arr to be length 3 or smaller function insertNum(arr, num) { if (!arr.length || num < arr[0]) { nums.unshift(num); } else if (num > arr[arr.length-1]) { arr.push(num); } else { arr[2] = arr[1]; arr[1] = num; } } var num, nums = []; if (data.length < 4) { return getMax(data); } for (var i=0, iLen=data.length; i<iLen; i++) { num = data[i]; // Insert first 3 values sorted if (nums.length < 3) { insertNum(nums, num); // If num is smaller than largest value in nums, // remove move largest and insert num } else if (num < nums[2]){ nums.splice(-1); insertNum(nums, num); } } // Return highest (last) value in nums return nums[2]; } var data = [3,2,55,-10,-55,5,3,2,1,-5,33,9,-1,4,5]; console.log(get3rdSmallest([-4,-22])); // -4 console.log(get3rdSmallest([4,0,1,7,6])); // 4 console.log(get3rdSmallest(data)); // -5