LeetCode 349. 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.