349. Intersection of Two Arrays

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

Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2].

Note:

Each element in the result must be unique.
The result can be in any order.

找两个数组相交（重复）的部分。用hashset存下第一个数组，遍历第二个数组，如果在set中含有的元素，就存在另一个set中，最后返回所有元素。代码如下：

public class Solution {    public int[] intersection(int[] nums1, int[] nums2) {        HashSet<Integer> nums1hs = new HashSet<Integer>();        HashSet<Integer> nums2hs = new HashSet<Integer>();        for (int i = 0; i < nums1.length; i++) {            nums1hs.add(nums1[i]);        }        for (int j = 0; j < nums2.length; j++) {            if (nums1hs.contains(nums2[j])) {                nums2hs.add(nums2[j]);            }        }        int[] result = new int[nums2hs.size()];        int index = 0;        for( Integer num : nums2hs ) {            result[index++] = num;         }        return result;    }}

还有另外两种解法，用sort 和binary search 的思想：

Sort both arrays, use two pointers

Time complexity: O(nlogn)

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 {                set.add(nums1[i]);                i++;                j++;            }        }        int[] result = new int[set.size()];        int k = 0;        for (Integer num : set) {            result[k++] = num;        }        return result;    }}

Binary search

Time complexity: O(nlogn)

public class Solution {    public int[] intersection(int[] nums1, int[] nums2) {        Set<Integer> set = new HashSet<>();        Arrays.sort(nums2);        for (Integer num : nums1) {            if (binarySearch(nums2, num)) {                set.add(num);            }        }        int i = 0;        int[] result = new int[set.size()];        for (Integer num : set) {            result[i++] = num;        }        return result;    }        public boolean binarySearch(int[] nums, int target) {        int low = 0;        int high = nums.length - 1;        while (low <= high) {            int mid = low + (high - low) / 2;            if (nums[mid] == target) {                return true;            }            if (nums[mid] > target) {                high = mid - 1;            } else {                low = mid + 1;            }        }        return false;    }}

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