给定两个数组，编写一个函数来计算它们的交集。

huangapple
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英文:

Given two arrays, write a function to compute their intersection

问题

  1.     class Solution {
  2.         public int[] intersect(int[] nums1, int[] nums2) {
  3.             HashSet<Integer> set1 = new HashSet<Integer>();
  4.             for (Integer n : nums1)
  5.                 set1.add(n);
  6.             HashSet<Integer> set2 = new HashSet<Integer>();
  7.             for (Integer n : nums2)
  8.                 set2.add(n);
  10.             set1.retainAll(set2);
  11.             int[] output = new int[set1.size()];
  12.             int idx = 0;
  13.             for (int s : set1)
  14.                 output[idx++] = s;
  15.             return output;
  16.         }
  17.     }
  19. output
  21.     Input:
  22.     [1,2,2,1]
  23.     [2,2]
  24.     Output:
  25.     [2]
  26.     Expected:
  27.     [2,2]
英文: 
  1. class Solution {
  2.     public int[] intersect(int[] nums1, int[] nums2) {
  3.         HashSet&lt;Integer&gt; set1= new HashSet&lt;Integer&gt;();
  4.         for (Integer n : nums1)
  5.             set1.add(n);
  6.         HashSet&lt;Integer&gt; set2 = new HashSet&lt;Integer&gt;();
  7.         for (Integer n : nums2)
  8.             set2.add(n);
  9.         
  10.         set1.retainAll(set2);
  11.         int[] output = new int[set1.size()];
  12.         int idx=0;
  13.         for(int s :set1)
  14.             output[idx++]=s;
  15.         return output;            
  16.     }
  17. }

output

  1. Input:
  2. [1,2,2,1]
  3. [2,2]
  4. Output:
  5. [2]
  6. Expected:
  7. [2,2]

答案1

得分: 0

集合不允许重复，因此如果您希望输出包含重复项，则它们不适用。

您可以使用Map<Integer, Integer>来计算第一个数组中每个元素的出现次数，然后与第二个数组进行比较：

  1. public int[] intersect(int[] nums1, int[] nums2) {
  2.     Map<Integer, Integer> map1 = new HashMap<>();
  3.     for (int n : nums1) {
  4.         if (map1.containsKey(n))
  5.             map1.put(n, map1.get(n) + 1);
  6.         else
  7.             map1.put(n, 1);
  8.     }
  9.     List<Integer> inter = new ArrayList<>();
  10.     for (int n : nums2) {
  11.         Integer count = map1.get(n);
  12.         if (count != null && count > 0) {
  13.             map1.put(n, count - 1);
  14.             inter.add(n);
  15.         }
  16.     }
  18.     int[] output = new int[inter.size()];
  19.     for (int i = 0; i < inter.size(); i++)
  20.         output[i] = inter.get(i);
  21.     return output;
  22. }
英文:

Sets don't allow duplicates, so they are not suitable if you expect the output to contain duplicates.

You can use a Map&lt;Integer,Integer&gt; to count the occurrences of each element of the first array, and then check them against the second array:

  1. public int[] intersect(int[] nums1, int[] nums2) {
  2.     Map&lt;Integer,Integer&gt; map1 = new HashMap&lt;&gt;();
  3.     for (int n : nums1) {
  4.         if (map1.containsKey(n))
  5.             map1.put(n,map1.get(n)+1);
  6.         else
  7.             map1.put(n,1);
  8.     }
  9.     List&lt;Integer&gt; inter = new ArrayList&lt;&gt;();
  10.     for (int n : nums2) {
  11.         Integer count = map1.get(n);
  12.         if (count != null &amp;&amp; count &gt; 0) {
  13.             map1.put(n,count-1);
  14.             inter.add(n);
  15.         }
  16.     }
  17.     
  18.     int[] output = new int[inter.size()];
  19.     for(int i = 0; i &lt; inter.size(); i++)
  20.         output[i]=inter.get(i);
  21.     return output;
  22.     
  23. }

huangapple
  • 本文由 发表于 2020年8月5日 16:29:49
  • 转载请务必保留本文链接：https://go.coder-hub.com/63261261.html
  • java
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