关键是使用索引来标记范围内是否存在某个数字。 这里我们知道从1到N。 时间复杂度O(n) 空间复杂度O(1)

后续问题: 这可以被修改为发现一个元素是否从差值为d的AP中缺失。其他变化可能包括从任何包含-ve数的随机数组中查找第一个缺失的+ve数。然后先对0左右的分区进行快速排序，然后对分区右侧的数组部分做此程序，做必要的修改。

public static void  missing(int [] arr){        
      for(int i=0; i< arr.length; i++){       
          if(arr[i]!=-1 && arr[i]<=arr.length){
              int idx=i;
              while(idx>=0 && idx<arr.length&& arr[idx]!=-1 ){
                   int temp =arr[idx];
                   // temp-1 because array index starts from 0, i.e a[0]=-1 is indicates that 1 is present in the array
                   arr[temp-1]=-1;
                   idx=temp-1;
              }
          }
      }
    }

在此之后，我们需要迭代数组，并检查是否a[i]!=-1，那么i+1就是缺失的数。当a[i]>N时，我们必须小心。

我们可以通过把数字本身和这些数字的平方相加来解Q2。

我们可以把问题简化为

k1 + k2 = x
k1^2 + k2^2 = y

其中x和y表示和低于期望值的程度。

代换给我们:

(x-k2)^2 + k2^2 = y

然后我们可以解出缺失的数。

正如@j_random_hacker所指出的，这与在O(n)个时间和O(1)个空间中寻找重复项非常相似，我的答案在这里也适用。

假设“袋子”由一个大小为N - k的基于1的数组a[]表示，我们可以在O(N)个时间和O(k)个额外空间内求解Qk。

首先，我们将数组A[]扩展k个元素，使它现在的大小为n，这是O(k)个额外空间。然后我们运行以下伪代码算法:

for i := n - k + 1 to n
    A[i] := A[1]
end for

for i := 1 to n - k
    while A[A[i]] != A[i] 
        swap(A[i], A[A[i]])
    end while
end for

for i := 1 to n
    if A[i] != i then 
        print i
    end if
end for

第一个循环初始化k个额外的条目，使其与数组中的第一个条目相同(这只是我们知道数组中已经存在的一个方便的值——在这一步之后，大小为N-k的初始数组中缺失的任何条目在扩展数组中仍然缺失)。

第二个循环排列扩展数组，如果元素x至少出现一次，那么其中一个元素将位于位置A[x]。

注意，尽管它有一个嵌套循环，但它仍然在O(N)时间内运行——只有当有一个i使a [i] != i时才会发生交换，并且每次交换设置至少一个元素使a [i] == i，而以前不是这样的。这意味着交换的总数(因此while循环体的执行总数)最多为N-1。

第三个循环打印数组i中没有被值i占用的索引——这意味着i一定是缺失的。

要解决缺少2(和3)个数字的问题，您可以修改quickselect，它平均在O(n)内运行，如果分区是就地完成的，则使用恒定内存。

Partition the set with respect to a random pivot p into partitions l, which contain numbers smaller than the pivot, and r, which contain numbers greater than the pivot. Determine which partitions the 2 missing numbers are in by comparing the pivot value to the size of each partition (p - 1 - count(l) = count of missing numbers in l and n - count(r) - p = count of missing numbers in r) a) If each partition is missing one number, then use the difference of sums approach to find each missing number. (1 + 2 + ... + (p-1)) - sum(l) = missing #1 and ((p+1) + (p+2) ... + n) - sum(r) = missing #2 b) If one partition is missing both numbers and the partition is empty, then the missing numbers are either (p-1,p-2) or (p+1,p+2) depending on which partition is missing the numbers. If one partition is missing 2 numbers but is not empty, then recurse onto that partiton.

由于只缺少2个数字，该算法总是丢弃至少一个分区，因此保持了O(n)个快速选择的平均时间复杂度。类似地，当缺少3个数字时，该算法也会在每次传递中丢弃至少一个分区(因为当缺少2个数字时，最多只有1个分区包含多个缺少的数字)。然而，我不确定当添加更多缺失的数字时，性能会下降多少。

下面是一个不使用就地分区的实现，所以这个例子不满足空间要求，但它确实说明了算法的步骤:

<?php

  $list = range(1,100);
  unset($list[3]);
  unset($list[31]);

  findMissing($list,1,100);

  function findMissing($list, $min, $max) {
    if(empty($list)) {
      print_r(range($min, $max));
      return;
    }

    $l = $r = [];
    $pivot = array_pop($list);

    foreach($list as $number) {
      if($number < $pivot) {
        $l[] = $number;
      }
      else {
        $r[] = $number;
      }
    }

    if(count($l) == $pivot - $min - 1) {
      // only 1 missing number use difference of sums
      print array_sum(range($min, $pivot-1)) - array_sum($l) . "\n";
    }
    else if(count($l) < $pivot - $min) {
      // more than 1 missing number, recurse
      findMissing($l, $min, $pivot-1);
    }

    if(count($r) == $max - $pivot - 1) {
      // only 1 missing number use difference of sums
      print array_sum(range($pivot + 1, $max)) - array_sum($r) . "\n";
    } else if(count($r) < $max - $pivot) {
      // mroe than 1 missing number recurse
      findMissing($r, $pivot+1, $max);
    }
  }

Demo

这是一个解决方案，它不依赖于复杂的数学，如sdcvvc /Dimitris Andreou的答案，不像caf和Colonel Panic那样改变输入数组，也不像Chris Lercher, JeremyP和许多其他人那样使用巨大的bitset。基本上，我从Svalorzen /Gilad Deutch关于Q2的想法开始，将其推广到常见情况Qk，并在Java中实现，以证明算法是有效的。

这个想法

假设我们有一个任意区间I其中我们只知道它包含至少一个缺失的数。在一次遍历输入数组后，只查看来自I的数字，我们可以获得I中缺失数字的S和Q。我们通过每次遇到来自I的数字时简单地减去I的长度(以获得Q)，并通过每次将I中所有数字的预计算总和减去遇到的数字(以获得S)来实现这一点。

Now we look at S and Q. If Q = 1, it means that then I contains only one of the missing numbers, and this number is clearly S. We mark I as finished (it is called "unambiguous" in the program) and leave it out from further consideration. On the other hand, if Q > 1, we can calculate the average A = S / Q of missing numbers contained in I. As all numbers are distinct, at least one of such numbers is strictly less than A and at least one is strictly greater than A. Now we split I in A into two smaller intervals each of which contains at least one missing number. Note that it doesn't matter to which of the intervals we assign A in case it is an integer.

We make the next array pass calculating S and Q for each of the intervals separately (but in the same pass) and after that mark intervals with Q = 1 and split intervals with Q > 1. We continue this process until there are no new "ambiguous" intervals, i.e. we have nothing to split because each interval contains exactly one missing number (and we always know this number because we know S). We start out from the sole "whole range" interval containing all possible numbers (like [1..N] in the question).

时空复杂性分析

在过程停止之前，我们需要通过的总次数p永远不会大于缺失数k。不等式p <= k可以被严格证明。另一方面，也有一个经验上限p < log2N + 3，这对于k的大值是有用的。我们需要对输入数组的每个数字进行二进制搜索，以确定它所属的区间。这给时间复杂度增加了log k乘数。

总的来说，时间复杂度为O(N᛫min(k, log N)᛫log k).注意，对于较大的k，这明显优于sdcvvc/Dimitris Andreou的方法，即O(N᛫k)。

对于它的工作，该算法需要O(k)个额外的空间来存储最多k个间隔，这明显优于“bitset”解决方案中的O(N)。

Java实现

下面是一个实现上述算法的Java类。它总是返回一个由缺失数字组成的有序数组。除此之外，它不需要缺少的数字计算k，因为它在第一次传递中计算k。整个数字范围由minNumber和maxNumber参数给出(例如，问题中的第一个例子是1和100)。

public class MissingNumbers {
    private static class Interval {
        boolean ambiguous = true;
        final int begin;
        int quantity;
        long sum;

        Interval(int begin, int end) { // begin inclusive, end exclusive
            this.begin = begin;
            quantity = end - begin;
            sum = quantity * ((long)end - 1 + begin) / 2;
        }

        void exclude(int x) {
            quantity--;
            sum -= x;
        }
    }

    public static int[] find(int minNumber, int maxNumber, NumberBag inputBag) {
        Interval full = new Interval(minNumber, ++maxNumber);
        for (inputBag.startOver(); inputBag.hasNext();)
            full.exclude(inputBag.next());
        int missingCount = full.quantity;
        if (missingCount == 0)
            return new int[0];
        Interval[] intervals = new Interval[missingCount];
        intervals[0] = full;
        int[] dividers = new int[missingCount];
        dividers[0] = minNumber;
        int intervalCount = 1;
        while (true) {
            int oldCount = intervalCount;
            for (int i = 0; i < oldCount; i++) {
                Interval itv = intervals[i];
                if (itv.ambiguous)
                    if (itv.quantity == 1) // number inside itv uniquely identified
                        itv.ambiguous = false;
                    else
                        intervalCount++; // itv will be split into two intervals
            }
            if (oldCount == intervalCount)
                break;
            int newIndex = intervalCount - 1;
            int end = maxNumber;
            for (int oldIndex = oldCount - 1; oldIndex >= 0; oldIndex--) {
                // newIndex always >= oldIndex
                Interval itv = intervals[oldIndex];
                int begin = itv.begin;
                if (itv.ambiguous) {
                    // split interval itv
                    // use floorDiv instead of / because input numbers can be negative
                    int mean = (int)Math.floorDiv(itv.sum, itv.quantity) + 1;
                    intervals[newIndex--] = new Interval(mean, end);
                    intervals[newIndex--] = new Interval(begin, mean);
                } else
                    intervals[newIndex--] = itv;
                end = begin;
            }
            for (int i = 0; i < intervalCount; i++)
                dividers[i] = intervals[i].begin;
            for (inputBag.startOver(); inputBag.hasNext();) {
                int x = inputBag.next();
                // find the interval to which x belongs
                int i = java.util.Arrays.binarySearch(dividers, 0, intervalCount, x);
                if (i < 0)
                    i = -i - 2;
                Interval itv = intervals[i];
                if (itv.ambiguous)
                    itv.exclude(x);
            }
        }
        assert intervalCount == missingCount;
        for (int i = 0; i < intervalCount; i++)
            dividers[i] = (int)intervals[i].sum;
        return dividers;
    }
}

For fairness, this class receives input in form of NumberBag objects. NumberBag doesn't allow array modification and random access and also counts how many times the array was requested for sequential traversing. It is also more suitable for large array testing than Iterable<Integer> because it avoids boxing of primitive int values and allows wrapping a part of a large int[] for a convenient test preparation. It is not hard to replace, if desired, NumberBag by int[] or Iterable<Integer> type in the find signature, by changing two for-loops in it into foreach ones.

import java.util.*;

public abstract class NumberBag {
    private int passCount;

    public void startOver() {
        passCount++;
    }

    public final int getPassCount() {
        return passCount;
    }

    public abstract boolean hasNext();

    public abstract int next();

    // A lightweight version of Iterable<Integer> to avoid boxing of int
    public static NumberBag fromArray(int[] base, int fromIndex, int toIndex) {
        return new NumberBag() {
            int index = toIndex;

            public void startOver() {
                super.startOver();
                index = fromIndex;
            }

            public boolean hasNext() {
                return index < toIndex;
            }

            public int next() {
                if (index >= toIndex)
                    throw new NoSuchElementException();
                return base[index++];
            }
        };
    }

    public static NumberBag fromArray(int[] base) {
        return fromArray(base, 0, base.length);
    }

    public static NumberBag fromIterable(Iterable<Integer> base) {
        return new NumberBag() {
            Iterator<Integer> it;

            public void startOver() {
                super.startOver();
                it = base.iterator();
            }

            public boolean hasNext() {
                return it.hasNext();
            }

            public int next() {
                return it.next();
            }
        };
    }
}

测试

下面给出了演示这些类用法的简单示例。

import java.util.*;

public class SimpleTest {
    public static void main(String[] args) {
        int[] input = { 7, 1, 4, 9, 6, 2 };
        NumberBag bag = NumberBag.fromArray(input);
        int[] output = MissingNumbers.find(1, 10, bag);
        System.out.format("Input: %s%nMissing numbers: %s%nPass count: %d%n",
                Arrays.toString(input), Arrays.toString(output), bag.getPassCount());

        List<Integer> inputList = new ArrayList<>();
        for (int i = 0; i < 10; i++)
            inputList.add(2 * i);
        Collections.shuffle(inputList);
        bag = NumberBag.fromIterable(inputList);
        output = MissingNumbers.find(0, 19, bag);
        System.out.format("%nInput: %s%nMissing numbers: %s%nPass count: %d%n",
                inputList, Arrays.toString(output), bag.getPassCount());

        // Sieve of Eratosthenes
        final int MAXN = 1_000;
        List<Integer> nonPrimes = new ArrayList<>();
        nonPrimes.add(1);
        int[] primes;
        int lastPrimeIndex = 0;
        while (true) {
            primes = MissingNumbers.find(1, MAXN, NumberBag.fromIterable(nonPrimes));
            int p = primes[lastPrimeIndex]; // guaranteed to be prime
            int q = p;
            for (int i = lastPrimeIndex++; i < primes.length; i++) {
                q = primes[i]; // not necessarily prime
                int pq = p * q;
                if (pq > MAXN)
                    break;
                nonPrimes.add(pq);
            }
            if (q == p)
                break;
        }
        System.out.format("%nSieve of Eratosthenes. %d primes up to %d found:%n",
                primes.length, MAXN);
        for (int i = 0; i < primes.length; i++)
            System.out.format(" %4d%s", primes[i], (i % 10) < 9 ? "" : "\n");
    }
}

大数组测试可以这样执行:

import java.util.*;

public class BatchTest {
    private static final Random rand = new Random();
    public static int MIN_NUMBER = 1;
    private final int minNumber = MIN_NUMBER;
    private final int numberCount;
    private final int[] numbers;
    private int missingCount;
    public long finderTime;

    public BatchTest(int numberCount) {
        this.numberCount = numberCount;
        numbers = new int[numberCount];
        for (int i = 0; i < numberCount; i++)
            numbers[i] = minNumber + i;
    }

    private int passBound() {
        int mBound = missingCount > 0 ? missingCount : 1;
        int nBound = 34 - Integer.numberOfLeadingZeros(numberCount - 1); // ceil(log_2(numberCount)) + 2
        return Math.min(mBound, nBound);
    }

    private void error(String cause) {
        throw new RuntimeException("Error on '" + missingCount + " from " + numberCount + "' test, " + cause);
    }

    // returns the number of times the input array was traversed in this test
    public int makeTest(int missingCount) {
        this.missingCount = missingCount;
        // numbers array is reused when numberCount stays the same,
        // just Fisher–Yates shuffle it for each test
        for (int i = numberCount - 1; i > 0; i--) {
            int j = rand.nextInt(i + 1);
            if (i != j) {
                int t = numbers[i];
                numbers[i] = numbers[j];
                numbers[j] = t;
            }
        }
        final int bagSize = numberCount - missingCount;
        NumberBag inputBag = NumberBag.fromArray(numbers, 0, bagSize);
        finderTime -= System.nanoTime();
        int[] found = MissingNumbers.find(minNumber, minNumber + numberCount - 1, inputBag);
        finderTime += System.nanoTime();
        if (inputBag.getPassCount() > passBound())
            error("too many passes (" + inputBag.getPassCount() + " while only " + passBound() + " allowed)");
        if (found.length != missingCount)
            error("wrong result length");
        int j = bagSize; // "missing" part beginning in numbers
        Arrays.sort(numbers, bagSize, numberCount);
        for (int i = 0; i < missingCount; i++)
            if (found[i] != numbers[j++])
                error("wrong result array, " + i + "-th element differs");
        return inputBag.getPassCount();
    }

    public static void strideCheck(int numberCount, int minMissing, int maxMissing, int step, int repeats) {
        BatchTest t = new BatchTest(numberCount);
        System.out.println("╠═══════════════════════╬═════════════════╬═════════════════╣");
        for (int missingCount = minMissing; missingCount <= maxMissing; missingCount += step) {
            int minPass = Integer.MAX_VALUE;
            int passSum = 0;
            int maxPass = 0;
            t.finderTime = 0;
            for (int j = 1; j <= repeats; j++) {
                int pCount = t.makeTest(missingCount);
                if (pCount < minPass)
                    minPass = pCount;
                passSum += pCount;
                if (pCount > maxPass)
                    maxPass = pCount;
            }
            System.out.format("║ %9d  %9d  ║  %2d  %5.2f  %2d  ║  %11.3f    ║%n", missingCount, numberCount, minPass,
                    (double)passSum / repeats, maxPass, t.finderTime * 1e-6 / repeats);
        }
    }

    public static void main(String[] args) {
        System.out.println("╔═══════════════════════╦═════════════════╦═════════════════╗");
        System.out.println("║      Number count     ║      Passes     ║  Average time   ║");
        System.out.println("║   missimg     total   ║  min  avg   max ║ per search (ms) ║");
        long time = System.nanoTime();
        strideCheck(100, 0, 100, 1, 20_000);
        strideCheck(100_000, 2, 99_998, 1_282, 15);
        MIN_NUMBER = -2_000_000_000;
        strideCheck(300_000_000, 1, 10, 1, 1);
        time = System.nanoTime() - time;
        System.out.println("╚═══════════════════════╩═════════════════╩═════════════════╝");
        System.out.format("%nSuccess. Total time: %.2f s.%n", time * 1e-9);
    }
}

在Ideone上试试吧

我让一个4岁的孩子来解决这个问题。他把数字分类，然后开始数。这有一个O(厨房地板)的空间要求，它的工作就像许多球丢失一样简单。