Leetcode: 349. Intersection of Two Arrays

Given two arrays, write a function to compute their intersection.

Example:
Given nums1 = [1, 2, 2, 1]nums2 = [2, 2], return [2].

Note:

  • Each element in the result must be unique.
  • The result can be in any order.

public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
    int count = 0;
    HashMap<Integer, Integer> twoSum1 = new HashMap<>();
    HashMap<Integer, Integer> twoSum2 = new HashMap<>();

    constructTwoPairMap(twoSum1, A, B);
    constructTwoPairMap(twoSum2, C, D);
    for (Integer pairSum : twoSum1.keySet()) {
        count += twoSum1.get(pairSum) * twoSum2.getOrDefault(pairSum * (-1), 0);
    }
    return count;
}

public void constructTwoPairMap(HashMap<Integer, Integer> map, int[] A, int[] B) {
    for (int i = 0; i < A.length; i++) {
        for (int j = 0; j < B.length; j++) {
            int sum = A[i] + B[j];
            map.put(sum, map.getOrDefault(sum, 0) + 1);
        }
    }
}

Partition Array

1. with target outside the array

    public int partitionArray(int[] nums, int k) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        int left = 0;
        int right = nums.length - 1;
        while (left <= right) {
            while (left <= right && nums[left] < k) {
                left++;
            }
            while (left <= right && nums[right] >= k) {
                right--;
            }
            if (left < right) {
                swap(nums, left, right);
                left++;
                right--;
            }
        }
        return left;
    }

    public void swap(int[] nums, int a, int b) {
        int tmp = nums[a];
        nums[a] = nums[b];
        nums[b] = tmp;
    }

2. with target with in the array(Quick select)

Implement rand7() using rand5()

Write a method to generate a random number between 1 and 7, given a method that generates a random number between 1 and 5 (i.e., implement rand7() using rand5()).

解答

rand5可以随机生成1,2,3,4,5;rand7可以随机生成1,2,3,4,5,6,7。 rand5并不能直接产生6,7,所以直接用rand5去实现函数rand7似乎不太好入手。 如果反过来呢?给你rand7,让你实现rand5,这个好实现吗?

一个非常直观的想法就是不断地调用rand7,直到它产生1到5之间的数,然后返回。 代码如下:

<code class="language-cpp" data-lang="cpp"><span class="kt">int</span> <span class="nf">Rand5</span><span class="p">(){</span>
    <span class="kt">int</span> <span class="n">x</span> <span class="o">=</span> <span class="o">~</span><span class="p">(</span><span class="mi">1</span><span class="o">&lt;&lt;</span><span class="mi">31</span><span class="p">);</span> <span class="c1">// max int</span>
    <span class="k">while</span><span class="p">(</span><span class="n">x</span> <span class="o">&gt;</span> <span class="mi">5</span><span class="p">)</span>
        <span class="n">x</span> <span class="o">=</span> <span class="n">Rand7</span><span class="p">();</span>
    <span class="k">return</span> <span class="n">x</span><span class="p">;</span>
<span class="p">}</span></code>

等等,这个函数可以等概率地产生1到5的数吗?首先,它确确实实只会返回1到5这几个数, 其次,对于这些数,都是由Rand7等概率产生的(1/7),没有对任何一个数有偏袒, 直觉告诉我们,Rand5就是等概率地产生1到5的。事实呢?让我们来计算一下, 产生1到5中的数的概率是不是1/5就OK了。比如说,让我们来计算一下Rand5生成1 的概率是多少。上面的函数中有个while循环,只要没生成1到5间的数就会一直执行下去。 因此,我们要的1可能是第一次调用Rand7时产生,也可能是第二次,第三次,…第n次。 第1次就生成1,概率是1/7;第2次生成1,说明第1次没生成1到5间的数而生成了6,7, 所以概率是(2/7)*(1/7),依次类推。生成1的概率计算如下:

<code>P(x=1)=1/7 + (2/7) * 1/7 + (2/7)^2 * 1/7 + (2/7)^3 * 1/7 + ...
      =1/7 * (1 + 2/7 + (2/7)^2 + ...) // 等比数列
      =1/7 * 1 / (1 - 2/7)
      =1/7 * 7/5
      =1/5
</code>

上述计算说明Rand5是等概率地生成1,2,3,4,5的(1/5的概率)。从上面的分析中, 我们可以得到一个一般的结论,如果a > b,那么一定可以用Randa去实现Randb。其中, Randa表示等概率生成1到a的函数,Randb表示等概率生成1到b的函数。代码如下:

<code class="language-cpp" data-lang="cpp"><span class="c1">// a &gt; b</span>
<span class="kt">int</span> <span class="nf">Randb</span><span class="p">(){</span>
    <span class="kt">int</span> <span class="n">x</span> <span class="o">=</span> <span class="o">~</span><span class="p">(</span><span class="mi">1</span><span class="o">&lt;&lt;</span><span class="mi">31</span><span class="p">);</span> <span class="c1">// max int</span>
    <span class="k">while</span><span class="p">(</span><span class="n">x</span> <span class="o">&gt;</span> <span class="n">b</span><span class="p">)</span>
        <span class="n">x</span> <span class="o">=</span> <span class="n">Randa</span><span class="p">();</span>
    <span class="k">return</span> <span class="n">x</span><span class="p">;</span>
<span class="p">}</span></code>

回到正题,现在题目要求我们要用Rand5来实现Rand7,只要我们将Rand5 映射到一个能产生更大随机数的Randa,其中a > 7,就可以套用上面的模板了。 这里要注意一点的是,你映射后的Randa一定是要满足等概率生成1到a的。比如,

<code>Rand5() + Rand5() - 1
</code>

上述代码可以生成1到9的数,但它们是等概率生成的吗?不是。生成1只有一种组合: 两个Rand5()都生成1时:(1, 1);而生成2有两种:(1, 2)和(2, 1);生成6更多。 它们的生成是不等概率的。那要怎样找到一个等概率生成数的组合呢?

我们先给出一个组合,再来进行分析。组合如下:

<code>5 * (Rand5() - 1) + Rand5()
</code>

Rand5产生1到5的数,减1就产生0到4的数,乘以5后可以产生的数是:0,5,10,15,20。 再加上第二个Rand5()产生的1,2,3,4,5。我们可以得到1到25, 而且每个数都只由一种组合得到,即上述代码可以等概率地生成1到25。OK, 到这基本上也就解决了。

套用上面的模板,我们可以得到如下代码:

<code class="language-cpp" data-lang="cpp"><span class="kt">int</span> <span class="nf">Rand7</span><span class="p">(){</span>
    <span class="kt">int</span> <span class="n">x</span> <span class="o">=</span> <span class="o">~</span><span class="p">(</span><span class="mi">1</span><span class="o">&lt;&lt;</span><span class="mi">31</span><span class="p">);</span> <span class="c1">// max int</span>
    <span class="k">while</span><span class="p">(</span><span class="n">x</span> <span class="o">&gt;</span> <span class="mi">7</span><span class="p">)</span>
        <span class="n">x</span> <span class="o">=</span> <span class="mi">5</span> <span class="o">*</span> <span class="p">(</span><span class="n">Rand5</span><span class="p">()</span> <span class="o">-</span> <span class="mi">1</span><span class="p">)</span> <span class="o">+</span> <span class="n">Rand5</span><span class="p">()</span> <span class="c1">// Rand25</span>
    <span class="k">return</span> <span class="n">x</span><span class="p">;</span>
<span class="p">}</span></code>

上面的代码有什么问题呢?可能while循环要进行很多次才能返回。 因为Rand25会产生1到25的数,而只有1到7时才跳出while循环, 生成大部分的数都舍弃掉了。这样的实现明显不好。我们应该让舍弃的数尽量少, 于是我们可以修改while中的判断条件,让x与最接近25且小于25的7的倍数相比。 于是判断条件可改为x > 21,于是x的取值就是1到21。 我们再通过取模运算把它映射到1-7即可。代码如下:

<code class="language-cpp" data-lang="cpp"><span class="kt">int</span> <span class="nf">Rand7</span><span class="p">(){</span>
    <span class="kt">int</span> <span class="n">x</span> <span class="o">=</span> <span class="o">~</span><span class="p">(</span><span class="mi">1</span><span class="o">&lt;&lt;</span><span class="mi">31</span><span class="p">);</span> <span class="c1">// max int</span>
    <span class="k">while</span><span class="p">(</span><span class="n">x</span> <span class="o">&gt;</span> <span class="mi">21</span><span class="p">)</span>
        <span class="n">x</span> <span class="o">=</span> <span class="mi">5</span> <span class="o">*</span> <span class="p">(</span><span class="n">Rand5</span><span class="p">()</span> <span class="o">-</span> <span class="mi">1</span><span class="p">)</span> <span class="o">+</span> <span class="n">Rand5</span><span class="p">()</span> <span class="c1">// Rand25</span>
    <span class="k">return</span> <span class="n">x</span><span class="o">%</span><span class="mi">7</span> <span class="o">+</span> <span class="mi">1</span><span class="p">;</span>
<span class="p">}</span></code>

这个实现就比上面的实现要好,并且可以保证等概率生成1到7的数。

让我们把这个问题泛化一下,从特殊到一般。现在我给你两个生成随机数的函数Randa, Randb。Randa和Randb分别产生1到a的随机数和1到b的随机数,a,b不相等 (相等就没必要做转换了)。现在让你用Randa实现Randb。

通过上文分析,我们可以得到步骤如下:

  1. 如果a > b,进入步骤2;否则构造Randa2 = a * (Randa – 1) + Randa, 表示生成1到a2 随机数的函数。如果a2 仍小于b,继教构造 Randa3 = a * (Randa2 – 1) + Randa…直到ak > b,这时我们得到Randak , 我们记为RandA。
  2. 步骤1中我们得到了RandA(可能是Randa或Randak ),其中A > b, 我们用下述代码构造Randb:
<code class="language-cpp" data-lang="cpp"><span class="c1">// A &gt; b</span>
<span class="kt">int</span> <span class="nf">Randb</span><span class="p">(){</span>
    <span class="kt">int</span> <span class="n">x</span> <span class="o">=</span> <span class="o">~</span><span class="p">(</span><span class="mi">1</span><span class="o">&lt;&lt;</span><span class="mi">31</span><span class="p">);</span> <span class="c1">// max int</span>
    <span class="k">while</span><span class="p">(</span><span class="n">x</span> <span class="o">&gt;</span> <span class="n">b</span><span class="o">*</span><span class="p">(</span><span class="n">A</span><span class="o">/</span><span class="n">b</span><span class="p">))</span> <span class="c1">// b*(A/b)表示最接近A且小于A的b的倍数</span>
        <span class="n">x</span> <span class="o">=</span> <span class="n">RandA</span><span class="p">();</span>
    <span class="k">return</span> <span class="n">x</span><span class="o">%</span><span class="n">b</span> <span class="o">+</span> <span class="mi">1</span><span class="p">;</span>
<span class="p">}</span></code>

从上面一系列的分析可以发现,如果给你两个生成随机数的函数Randa和Randb, 你可以通过以下方式轻松构造Randab,生成1到a*b的随机数。

<code>Randab = b * (Randa - 1) + Randb
Randab = a * (Randb - 1) + Randa
</code>

如果再一般化一下,我们还可以把问题变成:给你一个随机生成a到b的函数, 用它去实现一个随机生成c到d的函数。有兴趣的同学可以思考一下,这里不再讨论。

全书题解目录:

Cracking the coding interview–问题与解答

 

reference: http://www.hawstein.com/posts/19.10.html

Longest Consecutive Sequence

Longest Consecutive Sequence

Given an unsorted array of integers, find the length of the longest consecutive elements sequence.

Example

Given [100, 4, 200, 1, 3, 2],
The longest consecutive elements sequence is [1, 2, 3, 4]. Return its length: 4.

Clarification

Your algorithm should run in O(n) complexity.

Solution: 建一个hash表 对每一个num[i]上下浮动寻找num[i]+1, num[i]+2, …, 和num[i]-1, num[i]-2, …是不是在表里 如果找到的话就把该数从hash表里删除避免重复查找,所以建表时间是O(n) 每个数只会被访问一次,所以时间复杂度是O(n).

public int longestConsecutive(int[] num) {
    Set<Integer> numSet = new HashSet<>();
    for (int i = 0; i < num.length; i++) {
        numSet.add(num[i]);
    }
    int maxLength = 0;
    for (int i = 0; i < num.length; i++) {
        if (numSet.contains(num[i])) {
            int length = 1;
            int curr = num[i] - 1 ;
            while (numSet.contains(curr)) {
                numSet.remove(curr);
                length++;
                curr--;
            }
            curr = num[i] + 1;
            while (numSet.contains(curr)) {
                numSet.remove(curr);
                length++;
                curr++;
            }
            if (length > maxLength) {
                maxLength = length;
            }
        }
    }
    return maxLength;
}