LeetCode-Intersection of Two Arrays II

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最新推荐文章于 2024-03-31 22:07:26 发布

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分类专栏： LeetCode 文章标签： LeetCode Java Hash Table

本文链接：https://blog.youkuaiyun.com/qq_24133491/article/details/87597929

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本文介绍了一种求解两个数组交集的算法，通过使用哈希表存储元素及其出现次数，实现了对两个数组交集的有效计算，包括重复元素。文章提供了Java代码示例，展示了如何在遍历第二个数组时更新哈希表并确定交集元素。

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

Example 1:

Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2,2]

Example 2:

Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [4,9]

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?

题意：求两个数组的交集，可能包括重复的元素；

解法：考虑利用哈希表来实现，我们首先将其中一个数组的所有元素按照<元素，出现次数>来存储，之后遍历另外一个数组，每次执行以下的操作

如果哈希表中此元素的次数大于0，则当前元素为交集的元素；
如果哈希表中此元素的次数小于或等于0，则交集中不存在此元素或两个数组中此元素的个数不相同；

Java

class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        Map<Integer, Integer> frequency = new HashMap<>();
        List<Integer> res = new ArrayList<>();
        int cnt = 0;
        for (int num: nums1) {
            frequency.put(num, frequency.getOrDefault(num, 0) + 1);
        }
        for (int num: nums2) {
            frequency.put(num, frequency.getOrDefault(num, 0) - 1);
            if (frequency.get(num) >= 0) {
                res.add(num);
            }
        }
        return res.stream().mapToInt(Integer::valueOf).toArray();
    }
}