【Sort】350. Intersection of Two Arrays II
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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?
解答:
被分类到了Sort的题目,看标签和Binary Search, Hash Table, Two Pointers都有关。
刚开始学Hash Table,解答来自Discuss:
m: nums1.size n: nums2.size
Hash table solution:
Time: O(m + n) Space: O(m + n)
class Solution {public: vector<int> intersect(vector<int>& nums1, vector<int>& nums2) { unordered_map<int, int> dict; vector<int> 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; }};
Hash table solution2:
Time: O(m + n) Space: O(m)
class Solution {public: vector<int> intersect(vector<int>& nums1, vector<int>& nums2) { unordered_map<int, int> dict; vector<int> res; for(int i = 0; i < (int)nums1.size(); i++) dict[nums1[i]]++; for(int i = 0; i < (int)nums2.size(); i++) if(dict.find(nums2[i]) != dict.end() && --dict[nums2[i]] >= 0) res.push_back(nums2[i]); return res; }};
Sort and two pointers Solution:
Time: O(max(m, n) log(max(m, n))) Space: O(m + n)
class Solution {public: vector<int> intersect(vector<int>& nums1, vector<int>& 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<int> 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; }};
还有一种解答用了set_intersection():class Solution {public: vector<int> intersect(vector<int>& nums1, vector<int>& nums2) { sort(nums1.begin(), nums1.end()); sort(nums2.begin(), nums2.end()); nums1.erase(set_intersection(nums1.begin(), nums1.end(), nums2.begin(), nums2.end(), nums1.begin()), nums1.end());//利用了set_intersection(),是std::set return nums1; }};
这里补一下set_intersection这个函数,下面的代码描述了它的行为:
template <class InputIterator1, class InputIterator2, class OutputIterator> OutputIterator set_intersection (InputIterator1 first1, InputIterator1 last1, InputIterator2 first2, InputIterator2 last2, OutputIterator result){ while (first1!=last1 && first2!=last2) { if (*first1<*first2) ++first1; else if (*first2<*first1) ++first2; else { *result = *first1; ++result; ++first1; ++first2; } } return result;}
可见,最后第5个参数,也就是这里的nums1.begin()指向的是[2,2,0,0,0,0]的第一个0。
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