0350. Intersection of Two Arrays II (E)

Intersection of Two Arrays II (E)

题目

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?

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题意

取两数组的交集，包含所有重复元素

思路

普通做法直接用HashMap处理即可。

排序情况下用类似于归并排序的方法处理。

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代码实现

Java

Hash

class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        List list = new ArrayList<>();
        Map record = new HashMap<>();
        for (int num : nums1) {
            record.putIfAbsent(num, 0);
            record.put(num, record.get(num) + 1);
        }

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

        int[] res = new int[list.size()];
        int i = 0;
        for (int num : list) {
            res[i++] = num;
        }

        return res;
    }
}

排序

class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        List list = new ArrayList<>();
        Arrays.sort(nums1);
        Arrays.sort(nums2);
        int p1 = 0, p2 = 0;
        while (p1 < nums1.length && p2 < nums2.length) {
            if (nums1[p1] > nums2[p2]) {
                p2++;
            } else if (nums1[p1] < nums2[p2]) {
                p1++;
            } else {
                list.add(nums1[p1]);
                p1++;
                p2++;
            }
        }

        int[] res = new int[list.size()];
        int i = 0;
        for (int num : list) {
            res[i++] = num;
        }

        return res;
    }
}