LeetCode 350. Intersection of Two Arrays II

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

题目：

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?

题解：

可以使用双指针.

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

AC Java:

public class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        Arrays.sort(nums1);
        Arrays.sort(nums2);
        int i = 0;
        int j = 0;
        List<Integer> res = new ArrayList<Integer>();
        while(i<nums1.length && j<nums2.length){
            if(nums1[i] < nums2[j]){
                i++;
            }else if(nums1[i] > nums2[j]){
                j++;
            }else{
                res.add(nums1[i]);
                i++;
                j++;
            }
        }
        
        int [] resArr = new int[res.size()];
        int k = 0;
        for(int num : res){
            resArr[k++] = num;
        }
        return resArr;
    }
}

AC C++:

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        vector<int> res;
        sort(nums1.begin(), nums1.end());
        sort(nums2.begin(), nums2.end());
        int len1 = nums1.size();
        int len2 = nums2.size();
        int i = 0;
        int j = 0;
        while(i < len1 && j < len2){
            if(nums1[i] < nums2[j]){
                i++;
            }else if(nums1[i] > nums2[j]){
                j++;
            }else{
                res.push_back(nums1[i]);
                i++;
                j++;
            }
        }

        return res;
    }
};

Could use HashMap as well.

Time Complexity: O(m + n).

Space: O(Math.min(m, n)).

AC Java:

class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        if(nums1 == null || nums2 == null){
            return new int[0];
        }
        
        HashMap<Integer, Integer> hm = new HashMap<>();
        for(int num : nums1){
            hm.put(num, hm.getOrDefault(num, 0) + 1);
        }
        
        List<Integer> res = new ArrayList<>();
        for(int num : nums2){
            if(hm.containsKey(num)){
                res.add(num);
                if(hm.get(num) == 1){
                    hm.remove(num);
                }else{
                    hm.put(num, hm.get(num) - 1);
                }
            }
        }
        
        int [] resArr = new int[res.size()];
        int i = 0;
        for(int num : res){
            resArr[i++] = num;
        }
        
        return resArr;
    }
}

AC C++:

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        vector<int> res;
        unordered_map<int, int> map;
        for(int num : nums1){
            map[num]++;
        }

        for(int num : nums2){
            if(map[num] > 0){
                res.push_back(num);
                map[num]--;
            }
        }

        return res;
    }
};

Follow up 2 nums1 length is smaller. 用双指针先sort两个array明显没有利用到num1.length小的特性. 若是用HashMap来记录num1每个element出现频率再iterate nums2, 那么Time Complexity: O(m + n), m = nums1.length, n = num2.length. Space: O(m).

或者sort nums1 再对每一个num2的element在 sorted nums1上做 binary search. Time Complexity: O(mlogm + nlogm). Space: O(1).

由此可见，当m很小时，用HashMap和binary search就是time和space的trade off.

 

Follow up 3 nums2 is sorted but too big for memory. I/O的操作很贵，所以首先想到的是避免I/O的次数。

若是nums1可以全部load到memory上, 先sort nums1再把nums2从小到大分开load到memory来.

　　if load进来这一段最大值, 也就是最后一个值<nums1[0] 或者 load进来这一段最小值, 也就是第一个值>nums1[nums1.length-1]可以直接跳过, load下一段. 　　else load进来这一段 和 sorted nums1做双指针.

若是nums1也太大了，就先external sort nums1, 在分开load进来nums1一段和nums2一段做双指针.

类似Find Common Characters, Intersection of Two Arrays.