Given two arrays, write a function to compute their intersection.
Example:
Given nums1 = [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?
解法一:hashmap
public class Solution { public int[] intersect(int[] nums1, int[] nums2) { HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(); ArrayList<Integer> result = new ArrayList<Integer>(); for(int i = 0; i < nums1.length; i++) { if(map.containsKey(nums1[i])) map.put(nums1[i], map.get(nums1[i])+1); else map.put(nums1[i], 1); } for(int i = 0; i < nums2.length; i++) { if(map.containsKey(nums2[i]) && map.get(nums2[i]) > 0) { result.add(nums2[i]); map.put(nums2[i], map.get(nums2[i])-1); } } int[] r = new int[result.size()]; for(int i = 0; i < result.size(); i++) { r[i] = result.get(i); } return r; } }
解法二:
Sort both arrays, use two pointers
Time complexity: O(nlogn)
public class Solution { public int[] intersect(int[] nums1, int[] nums2) { List<Integer> res = new ArrayList<Integer>(); Arrays.sort(nums1); Arrays.sort(nums2); for(int i = 0, j = 0; i< nums1.length && j<nums2.length;){ if(nums1[i]<nums2[j]){ i++; } else if(nums1[i] == nums2[j]){ res.add(nums1[i]); i++; j++; } else{ j++; } } int[] result = new int[res.size()]; for(int i = 0; i<res.size();i++){ result[i] = res.get(i); } return result; } }
时间: 2024-12-18 06:25:20