# JAVA O(n) solution with HashMap & O(nlogn) solution with merge sort idea

• Here come two solutions based on Java.

1. Using HashMap to record and find out the duplicate part, O(n) time and O(n) space.
``````public int[] intersect(int[] nums1, int[] nums2) {
Map<Integer, Integer> map = new HashMap<>();
List<Integer> temp = new ArrayList<Integer>();
for (int num : nums1)
map.put(num, map.getOrDefault(num, 0)+1);
for (int num : nums2){
if (map.containsKey(num) && map.get(num) > 0){
map.put(num, map.get(num)-1);
}
}
int[] res = new int[temp.size()];
for (int i = 0; i < temp.size(); i++)
res[i] = temp.get(i);
return res;
}
``````
1. Using Merge Sort idea, O(nlogn) time and O(1) space.
``````public int[] intersect(int[] nums1, int[] nums2) {
Arrays.sort(nums1);
Arrays.sort(nums2);
List<Integer> temp = new ArrayList<Integer>();
for (int i = 0, j = 0; i < nums1.length && j < nums2.length;){
if (nums1[i] < nums2[j]) i++;
else if (nums1[i] > nums2[j]) j++;
else {
i++; j++;
}
}
int[] res = new int[temp.size()];
for (int i = 0; i < temp.size(); i++)
res[i] = temp.get(i);
return res;
}``````

• good solution

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