350. Intersection of Two Arrays II

發表於 | 分類於

problem

solution

option 1 - hash table

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class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        unordered_map<int,int>mp;
        vector<int> ret;
        for(int n:nums1) mp[n]++;
        for(int n:nums2){
            if(mp.find(n)!=mp.end() && mp[n]!=0){
                mp[n]--;
                ret.push_back(n);
            }
        }
        return ret;
    }
};

option 2 - improve hash table

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class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        vector<int> freq(10001,0), ret;
        for(int n:nums1) freq[n]++;
        for(int n:nums2){
            if(freq[n]>0){
                ret.push_back(n);
                freq[n]--;
            }
        }
        return ret;
        
    }
};

option 3 - sorting

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class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        vector<int> ret;
        sort(nums1.begin(), nums1.end());
        sort(nums2.begin(), nums2.end());
        int l=0,r=0, n=nums1.size(), m = nums2.size();
        while(l<n && r<m){
            if(nums1[l] == nums2[r]) {
                ret.push_back(nums1[l]);
                // while(l+1<n && nums1[l] == nums1[l+1]) l++;
                // while(r+1<m && nums2[r] == nums2[r+1]) r++;
                l++;
                r++;
            }
            else{
                if(nums1[l]<nums2[r]) l++;
                else r++;
            }
        }
        return ret;
    }
};

analysis

  • option 1
    • time complexity O(nlogn)
    • space complexity O(n)
  • option 1
    • time complexity O(n)
    • space complexity O(1)
  • option 3 - sorting
    • time complexity O(nlogn)
    • space complexity O(1)