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# 350 Intersection of Two Arrays II

## 350. [Intersection of Two Arrays II](https://leetcode.com/problems/intersection-of-two-arrays-ii/description/)

## 1. Question

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

**Example:**\
Givennums1=`[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?

## 2. Implementation

**(1) Two Pointers**

```java
class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        Arrays.sort(nums1);
        Arrays.sort(nums2);

        List<Integer> list = new ArrayList<>();

        int i = 0, j = 0;

        while (i < nums1.length && j < nums2.length) {
            if (nums1[i] < nums2[j]) {
                ++i;
            }
            else if (nums1[i] > nums2[j]) {
                ++j;
            }
            else {
                list.add(nums1[i]);
                ++i;
                ++j;
            }
        }

        int[] res = new int[list.size()];
        int index = 0;

        for (int e : list) {
            res[index++] = e;
        }
        return res;
    }
}
```

**(2) Binary Search**

```java
class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        if (nums1 == null || nums1.length == 0 || nums2 == null || nums2.length == 0) {
            return new int[0];
        }

        if (nums1.length < nums2.length) {
            return getIntersection(nums1, nums2);
        }
        else {
            return getIntersection(nums2, nums1);
        }
    }

    public int[] getIntersection(int[] nums1, int[] nums2) {
        List<Integer> list = new ArrayList<>();
        List<Integer> intersection = new ArrayList<>();

        Arrays.sort(nums2);

        for (int num : nums2) {
            list.add(num);
        }

        for (int i = 0; i < nums1.length; i++) {
            int searchIndex = binarySearch(list, nums1[i]);

            if (searchIndex != -1) {
                intersection.add(nums1[i]);
                list.remove(searchIndex);
            }
        }

        int[] res = new int[intersection.size()];
        int index = 0;

        for (int e : intersection) {
            res[index++] = e;
        }
        return res;
    }

    public int binarySearch(List<Integer> list, int target) {
        int start = 0, end = list.size() - 1, mid = 0;

        while (start <= end) {
            mid = start + (end - start) / 2;

            if (list.get(mid) == target) {
                return mid;
            }
            else if (list.get(mid) < target) {
                start = mid + 1;
            }
            else {
                end = mid - 1;
            }
        }
        return -1;
    }
}
```

## 3. Time & Space Complexity

**Two Pointers:** 时间复杂度O(nlogn + mlogm),n为nums1的长度, m为nums2的长度，空间复杂度O(Math.min(m,n))

**Binary Search**: 时间复杂度O(nlogn + mlogn), m为num1的长度, n为nums2的长度，空间复杂度O(Math.min(m,n))