Description

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?

My solution

延续leetcode 344的思路,少做修改即可, 但是复杂度仍旧很高啊...

class Solution {public:    vector
    
      intersect(vector
     
       &nums1, vector
      
        &nums2) {        vector
       
         res;        sort(nums1.begin(), nums1.end());        sort(nums2.begin(), nums2.end());        int i = 0, j = 0;        while (i < nums1.size() && j < nums2.size()) {            if (nums1[i] < nums2[j]) ++i;            else if (nums1[i] > nums2[j]) ++j;            else {                while (nums1[i] == nums2[j] && i < nums1.size() && j < nums2.size()) {                    res.push_back(nums1[i]);                    ++i;                    ++j;                }                while (nums1[i] == nums1[i - 1]) ++i;                while (nums2[j] == nums2[j - 1]) ++j;            }        }        return res;    }};

Discuss

sort

Time: O(max(m, n) log(max(m, n)))

但我又不得不说, 同样是sort方式, 我的方案太丑!!

discuss中同样方法的代码如下:

class Solution {public:    vector
    
      intersect(vector
     
      & nums1, vector
      
       & nums2) {        sort(nums1.begin(), nums1.end());        sort(nums2.begin(), nums2.end());        int n1 = (int)nums1.size(), n2 = (int)nums2.size();        int i1 = 0, i2 = 0;        vector
       
         res;        while(i1 < n1 && i2 < n2){            if(nums1[i1] == nums2[i2]) {                res.push_back(nums1[i1]);                i1++;                i2++;            }            else if(nums1[i1] > nums2[i2]){                i2++;            }            else{                i1++;            }        }        return res;    }};

注意, 只用了一次while! 逻辑很清晰!

Hash table solution

Time: O(m + n) Space: O(m + n)

class Solution {public:    vector
    
      intersect(vector
     
      & nums1, vector
      
       & nums2) {        unordered_map
       
         dict;        vector
        
          res;        for(int i = 0; i < (int)nums1.size(); i++) dict[nums1[i]]++;        for(int i = 0; i < (int)nums2.size(); i++)            if(--dict[nums2[i]] >= 0) res.push_back(nums2[i]);        return res;    }};