Find first missing number in a sequence of numbers

问题

I am trying to figure out how to find the first missing number of a sequence of numbers like this (1,2,3,5,6,9,10,15)

I want to put the first missing number, #4, into an variable for later use but don't know how to do so?

I have tried this but this only gives me the last number:

var mynumbers=new Array(1,2,3,6,9,10);
for(var i = 1; i < 32; i++) {
    if(mynumbers[i] - mynumbers[i-1] != 1) {
        alert("First missing number id: "+mynumbers[i]);
        break;
    }
}

First of all it gives me the first number after an "hole" in the numbersequence, secondly it continues to alert all numbers comming after an "hole" if I don't insert an break. I only want the first missing number of an numbersequence from 1 - 32. How do i do so?

Hoping for help and thanks in advance ;-)

回答1:

How about this

var mynumbers = new Array(1,2,3,6,9,10);
var missing;

for(var i=1;i<=32;i++)
{    
   if(mynumbers[i-1] != i){
        missing = i;
        alert(missing);
        break;
   }
}

回答2:

The O(n) solutions are easy , but this is a common interview question and often we look for O(log n) time solution. Here is the javascript code. It's basically a modified binary search.

function misingNumInSeq(source, min = 0, max = source.length - 1){
    if(min >= max){
        return min + 1;
    }
    let pivot = Math.floor((min + max)/2);
    // problem is in right side. Only look at right sub array
    if(source[pivot] === pivot + 1){
        return misingNumInSeq(source, pivot + 1, max);
    } else {
        return misingNumInSeq(source, min , pivot);
    }
}

Output

misingNumInSeq([1,2,3,5,6,9,10,15])
4

回答3:

By if(mynumbers[i] - mynumbers[i-1] != 1), you mean to say the series will always be incrementing by 1?

var missing = (function (arr) {
    var i;
    for (i = 0; i < arr.length; ++i) {
        if (i + arr[0] !== arr[i]) return i + arr[0];
    }
    if (i < 32)            // if none missing inside array and not yet 32nd
        return i + arr[0]; // return next
}([1,2,3,6,9,10])); // 4
alert(missing);

回答4:

You're going to need the break no matter what. That's what it's there for; to stop the loop from continuing on to the end. And you should use the length of the array instead of hardcoding 32 as the end condition, because your numbers only go up to 32, but there are possibly holes in the list so there will not be 32 elements in the array.

Since you know that each element should be 1 more than the previous element, then the number in the hole is clearly mynumbers[i - 1] + 1.

var mynumbers = new Array(1,2,3,6,9,10);
for(var i = 1; i < mynumbers.length; i++) {
    if(mynumbers[i] - mynumbers[i-1] != 1) {
        alert("First missing number id: " + (mynumbers[i - 1] + 1));
        break;
    }
}

EDIT: This only holds true for the missing number not being 1. To catch that, you will need to check if (mynumbers[0] != 1)

回答5:

Edit:

function findFirstMissing(array) {
    for (var i = 0; i < array.length; i++) {
        if (i+1 !== array[i]) {
            return i+1;
        }
    }
}

function findFirstMissing(array) {
    for (var i = 0; i < array.length; i++) {
        if (array[i+1] - array[i] !== 1) {
            return array[i] + 1;
        }
    }
}

If you do it this way then storing it in a variable is easy:

var missing = findFirstMissing(array);

回答6:

const firstNonConsecutive = arr => arr.find((el, i, arr) => (arr[i] - arr[i-1]) !== 1 && i !== 0)

this solution work for an array of positive numbers.

回答7:

A solution using array.reduce to find the first positive missing integer.

function solution(A) {
  return [...A].sort().reduce((acc, curr, i, arr) => {
    if (acc > curr) {
      arr.splice(1);
      return acc;
    }
    else if (arr[i + 1] - curr > 1 || arr.length === i + 1) {
      arr.splice(1);
      return curr + 1;
    }
    return acc;
  }, 1);
}

And here are few test cases:

console.log('solution([1, 3, 6, 4, 1, 2])',  solution([1, 3, 6, 4, 1, 2]) === 5)
console.log('solution([1, 3, 2, 8, 4])', solution([1, 3, 2, 8, 4]) === 5)
console.log('solution([1])', solution([1]) === 2)
console.log('solution([-1])', solution([-1]) === 1)
console.log('solution([0])', solution([0]) === 1)
console.log('solution([-1, -4, -5, -6, -190343])', solution([-1, -4, -5, -6, -190343]) === 1)

回答8:

Sometimes you just want simple if you know it's a small array:

let numbers = [1,2,3,6,9,10]
let m = 0

for (const i of numbers) if (i > ++m) break

console.log(m) // 4

Works if you remove 1 from start of array:

numbers = [2,3,6,9,10]
m = 0

for (const i of numbers) if (i > ++m) break

console.log(m) // 1

If the array can be contiguous, and if so you want the next highest number, then:

numbers = [1,2,3,4,5,6,7,8,9]
m = 0

for (const i of numbers) if (i > ++m) break
if (m == Math.max(...numbers)) m++

console.log(m) // 10

Short and sweet!

回答9:

I think this is the simplest and optimum form of just a 2-step solution.

I think no better solution can be possible for this problem than this one.

This code uses minimum no. of variables, loops, conditionals, built-in functions and all the shitty, sloppy, unnecessary code.

This code can handle array of any length.

var mynumbers = new Array(76,77,78,79,80,81,82,83,84,125);

if(mynumbers.length > 1) {

  for(var i=0; i<=mynumbers.length-1; i++) {

    if(mynumbers[i+1] - 1 !== mynumbers[i]) { 

      alert("First Missing Term is : "+parseInt(mynumbers[i]+1));
      break;
    }

  }

}

回答10:

for(var i = 1; i < mynumbers.length; i++) {
    if(mynumbers[i] - mynumbers[i-1] != 1) {
        alert("First missing number id: "+mynumbers[i-1]+1);
        i = mynumbers.length; // Replace the break
    }
}

If you want you can add an initial check : if (mynumbers[0] != 1) { ... }

来源：https://stackoverflow.com/questions/18853280/find-first-missing-number-in-a-sequence-of-numbers

标签

javascript

numbers

Sequence