README.md

Intersection of Two Arrays II

Description

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?

思路

使用 HashMap 遍历第一个数组，并记录数组元素出现的次数。 然后遍历第二个数组，如果数组元素在 HashMap 中，则把该元素添加到 List，同时 HashMap 该元素出现次数减一。 最后把 List 转成 int 数组即可。 时间复杂度 O(n)，空间复杂度 O(n)。

实现代码

class Solution {
  public int[] intersect(int[] nums1, int[] nums2) {
    Map<Integer, Integer> map = new HashMap<Integer, Integer>();
    List<Integer> list = new ArrayList<Integer>();

    for (int num : nums1) {
      if (map.containsKey(num)) {
        map.put(num, map.get(num) + 1);
      } else {
        map.put(num, 1);
      }
    }

    for (int num : nums2) {
      if (map.containsKey(num)) {
        if (map.get(num) > 0) {
          list.add(num);
          map.put(num, map.get(num) - 1);
        }
      }
    }

    int size = list.size();
    int result[] = new int[size];
    for (int i = 0; i < size; i++) {
      result[i] = list.get(i);
    }

    return result;
  }
}