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# 349.Intersection-of-Two-Arrays

## 题目描述

Given two arrays, write a function to compute their intersection.
Example 1:
Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2]
Example 2:
Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [9,4]
Note:
Each element in the result must be unique.
The result can be in any order.

## 代码

### Approach #1 Two Sets

class Solution {
public int[] intersection(int[] nums1, int[] nums2) {
HashSet<Integer> set1 = new HashSet<Integer>();
HashSet<Integer> set2 = new HashSet<Integer>();
if (set1.size() < set2.size()) {
return helper(set1, set2);
} else {
return helper(set2, set1);
}
}
private int[] helper(HashSet<Integer> set1, (HashSet<Integer> set2) {
int[] output = new int[set1.size()];
int idx = 0;
for (Integer s: set1) {
if (set2.contains(s)) {
output[idx++] = s;
}
}
return Arrays.copyOf(output, idx);
}
}

### Approach #2 Built-in Set Intersection

class Solution {
public int[] intersection(int[] nums1, int[] nums2) {
HashSet<Integer> set1 = new HashSet<Integer>();
for (Integer n : nums1) set1.add(n);
HashSet<Integer> set2 = new HashSet<Integer>();
for (Integer n : nums2) set2.add(n);
set1.retainAll(set2);
int [] output = new int[set1.size()];
int idx = 0;
for (int s : set1) output[idx++] = s;
return output;
}
}

### Approach #3 Two Pointer

public class Solution {
public int[] intersection(int[] nums1, int[] nums2) {
Set<Integer> set = new HashSet<>();
Arrays.sort(nums1);
Arrays.sort(nums2);
int i = 0;
int j = 0;
while (i < nums1.length && j < nums2.length) {
if (nums1[i] < nums2[j]) {
i++;
} else if (nums1[i] > nums2[j]) {
j++;
} else {