More:【目錄】LeetCode Java實現html
https://leetcode.com/problems/intersection-of-two-arrays-ii/
java
Given two arrays, write a function to compute their intersection.app
Example 1:post
Input: nums1 = [1,2,2,1], nums2 = [2,2] Output: [2,2]
Example 2:ui
Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4] Output: [4,9]
Note:spa
Follow up:code
Method 1: HashMaphtm
Method 2: Sorted arrays and two pointersblog
//Method 1: HashMap public int[] intersect(int[] nums1, int[] nums2) { HashMap<Integer,Integer> map = new HashMap<>(); ArrayList<Integer> list = new ArrayList<>(); for(int i : nums1){ if(map.containsKey(i)){ map.put(i, map.get(i)+1); }else{ map.put(i, 1); } } for(int i : nums2){ if(map.containsKey(i) && map.get(i)>0){ map.put(i, map.get(i)-1); list.add(i); } } int[] arr = new int[list.size()]; for(int n=0; n<list.size(); n++){ arr[n]=list.get(n); } return arr; } //Method 2: Sorted array & Two pointers public int[] intersect2(int[] nums1, int[] nums2) { Arrays.sort(nums1); Arrays.sort(nums2); ArrayList<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++]); j++; } } int[] arr = new int[list.size()]; for(int n=0; n<list.size(); n++){ arr[n]=list.get(n); } return arr; }
Method 1: ip
Time complexity : O(N)
Space complexity : O(N)
Method 2:
Time complexity : O(NlogN)
Space complexity : O(1)
More:【目錄】LeetCode Java實現