leetcode 350. Intersection of Two Arrays II 两个数组的交集

悠悠 2022-06-07 01:16 212阅读 0赞

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?

和上一道题[leetcode 349. Intersection of Two Arrays 数组的交集][leetcode 349. Intersection of Two Arrays] 做法一样，直接使用HashMap来做就可以了，注意这里存在重复的元素。

代码如下：

import java.util.ArrayList;
    import java.util.Arrays;
    import java.util.HashMap;
    import java.util.List;
    import java.util.Map;
    
    /*
     * 使用HashMap要不使用Set高明多了
     * */
    public class Solution 
    {
          public int[] intersectByHashMap(int[] nums1, int[] nums2) 
          {
                if(nums1==null || nums1.length<=0 || nums2==null || nums2.length<=0)
                    return new int[0];
    
                List<Integer> res=new ArrayList<Integer>();
                Map<Integer, Integer> map1=new HashMap<>();
                for(int i=0;i<nums1.length;i++)
                    map1.put(nums1[i], map1.getOrDefault(nums1[i], 0)+1);
    
                for(int i=0;i<nums2.length;i++)
                {
                    if(map1.containsKey(nums2[i]) && map1.get(nums2[i])>0)
                    {
                        res.add(nums2[i]);
                        map1.put(nums2[i], map1.get(nums2[i])-1);
                    }
                }
                int[] res1=new int[res.size()];
                for(int i=0;i<res1.length;i++)
                    res1[i]=res.get(i);
                return res1;
            }
    
             /*
              * 使用排序去做 
              * */
              public int[] intersect(int[] nums1, int[] nums2) 
              {
                    if(nums1==null || nums1.length<=0 || nums2==null || nums2.length<=0)
                        return new int[0];
    
                    List<Integer> res=new ArrayList<Integer>();
                    Arrays.sort(nums1);
                    Arrays.sort(nums2);
    
                    int i=0,j=0;
                    while(i<nums1.length && j<nums2.length)
                    {
                        if(nums1[i]==nums2[j])
                        {
                            res.add(nums1[i]);
                            i++;
                            j++;
                        }else if(nums1[i] > nums2[j])
                            j++;
                        else 
                            i++;
                    }
                    int[] res1=new int[res.size()];
                    for(int k=0;k<res1.length;k++)
                        res1[k]=res.get(k);
                    return res1;
                }
    }

下面是C++的做法，和上一道题一样直接使用map数据结构处理做查询即可，注意查询结束之后要删掉相关元素

不要要map。直接multiset即可

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <unordered_map>
    #include <set>
    #include <unordered_set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    #include <regex>
    #include <iomanip>
    #include <cstdlib>
    #include <ctime>
    
    using namespace std;
    
    
    
    class Solution 
    {
    public:
        vector<int> intersect(vector<int>& nums1, vector<int>& nums2) 
        {
            multiset<int> s;
            for (int i : nums1)
                s.insert(i);
    
            vector<int> res;
            for (int i : nums2)
            {
                if (s.find(i) != s.end())
                {
                    res.push_back(i);
                    s.erase(s.find(i));
                }
            }
            return res;
        }
    };

[leetcode 349. Intersection of Two Arrays]: http://blog.csdn.net/jackzhang_123/article/details/78151944