LeetCode Intersection of Two Arrays

原题链接在这里：https://leetcode.com/problems/intersection-of-two-arrays/

题目：

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.

题解：

用一个HashSet 来保存nums1的每个element.

再iterate nums2, 若HashSet contains nums2[i], 把nums2[i]加到res中，并把nums[i]从HashSet中remove掉.

Time Complexity: O(nums1.length + nums2.length). Space: O(nums1.length).

AC Java:

 public class Solution {
     public int[] intersection(int[] nums1, int[] nums2) {
         HashSet<Integer> nums1Hs = new HashSet<Integer>();
         for(int num : nums1){
             nums1Hs.add(num);
         }

         List<Integer> res = new ArrayList<Integer>();
         for(int num : nums2){
             if(nums1Hs.contains(num)){
                 res.add(num);
                 nums1Hs.remove(num);
             }
         }
         int [] resArr = new int[res.size()];
         int i = 0;
         for(int num : res){
             resArr[i++] = num;
         }
         return resArr;
     }
 }

也可以使用两个HashSet.

Time Complexity: O(nums1.length + nums2.length). Space: O(nums1.length).

AC Java:

 public class Solution {
     public int[] intersection(int[] nums1, int[] nums2) {
         HashSet<Integer> nums1Hs = new HashSet<Integer>();
         HashSet<Integer> intersectHs = new HashSet<Integer>();
         for(int num : nums1){
             nums1Hs.add(num);
         }
         for(int num : nums2){
             if(nums1Hs.contains(num)){
                 intersectHs.add(num);
             }
         }

         int [] res = new int[intersectHs.size()];
         int i = 0;
         for(int num : intersectHs){
             res[i++] = num;
         }
         return res;
     }
 }

Sort nums1 and nums2, 再用双指针 从头iterate两个sorted array.

Time Complexity: O(nlogn). Space: O(1).

AC Java:

 public class Solution {
     public int[] intersection(int[] nums1, int[] nums2) {
         Arrays.sort(nums1);
         Arrays.sort(nums2);
         HashSet<Integer> hs = new HashSet<Integer>();
         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{
                 hs.add(nums1[i]);
                 i++;
                 j++;
             }
         }

         int [] resArr = new int[hs.size()];
         int k = 0;
         for(int num : hs){
             resArr[k++] = num;
         }
         return resArr;
     }
 }

sort nums1, 然后nums2 array 每一个element在 sorted 上做binary search.

Time Complexity: O(mlogm + nlogm), m = nums1.length, n = nums2.length.

Space: O(resArr.length).

AC Java:

 public class Solution {
     public int[] intersection(int[] nums1, int[] nums2) {
         HashSet<Integer> res = new HashSet<Integer>();
         Arrays.sort(nums1);
         for(int num : nums2){
             if(binarySearch(nums1, num)){
                 res.add(num);
             }
         }

         int [] resArr = new int[res.size()];
         int i = 0;
         for(int num : res){
             resArr[i++] = num;
         }
         return resArr;
     }

     private 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){
                 low = mid+1;
             }else if(nums[mid] > target){
                 high = mid-1;
             }else{
                 return true;
             }
         }
         return false;
     }
 }

跟上Intersection of Two Arrays II, Intersection of Three Sorted Arrays.