350 Intersection of Two Arrays II

最新推荐文章于 2020-10-16 02:55:14 发布

原创最新推荐文章于 2020-10-16 02:55:14 发布 · 136 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 同时被 2 个专栏收录

13 篇文章

订阅专栏

Java

12 篇文章

订阅专栏

本文介绍了一种计算两个数组交集的方法，并提供了详细的实现步骤。针对不同情况，如已排序数组、大小不一的数组及存储在磁盘上的元素等，讨论了算法优化策略。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

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

Note:
Each element in the result should appear as many times as it shows in both arrays.
The result can be in any order.
Follow up:
What if the given array is already sorted? How would you optimize your algorithm?
What if nums1’s size is small compared to nums2’s size? Which algorithm is better?
What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?

class Solution {
  public int[] intersect(int[] nums1, int[] nums2) {
    Map<Integer, Integer> map1 = new HashMap<Integer, Integer>();
    Map<Integer, Integer> map2 = new HashMap<Integer, Integer>();

    // write numbers in nums1 in map1
    for (int n : nums1) {
      if (!map1.containsKey(n)) { map1.put(n, 1); }
      else { map1.put(n, 1 + map1.get(n)); }
    }
    // compute the count of result
    int count = 0;
    // if n in nums1 and nums2 put it in map2
    for (int n : nums2) {
      if (map1.containsKey(n)) {
        if (map2.containsKey(n)) {
          if (map1.get(n) <= map2.get(n)) { continue; }
          else { map2.put(n, 1 + map2.get(n)); count += 1;}
        } else {
          map2.put(n, 1);
          count += 1;
        }
      }
    }

    // write numbers in map2 to result
    int[] result = new int[count];
    count -= 1;
    for (Map.Entry<Integer, Integer> entry : map2.entrySet()) {
      for (int i = 0; i < entry.getValue(); i++) {
        result[count] = entry.getKey();
        count -= 1;
      }
    }
    return result;
  }
}