Given two arrays, write a function to compute their intersection.
Example 1:
Input: nums1 = [1,2,2,1], nums2 = [2,2] Output: [2,2]
Example 2:
Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4] Output: [4,9]
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?
這道題是之前那道
Intersection of Two Arrays 的拓展，不同之處在於這道題允許返回重復的數字，而且是盡可能多的返回，之前那道題是說有重復的數字只返回一個就行。那么這道題用 HashMap 來建立 nums1 中字符和其出現個數之間的映射, 然后遍歷 nums2 數組，如果當前字符在 HashMap 中的個數大於0，則將此字符加入結果 res 中，然后 HashMap 的對應值自減1，參見代碼如下：
解法一：
class Solution { public: vector<int> intersect(vector<int>& nums1, vector<int>& nums2) { unordered_map<int, int> m; vector<int> res; for (auto a : nums1) ++m[a]; for (auto a : nums2) { if (m[a]-- > 0) res.push_back(a); } return res; } };
再來看一種方法，這種方法先給兩個數組排序，然后用兩個指針分別指向兩個數組的起始位置，如果兩個指針指的數字相等，則存入結果中，兩個指針均自增1，如果第一個指針指的數字大，則第二個指針自增1，反之亦然，參見代碼如下：
解法二：
class Solution { public: vector<int> intersect(vector<int>& nums1, vector<int>& nums2) { vector<int> res; int i = 0, j = 0; sort(nums1.begin(), nums1.end()); sort(nums2.begin(), nums2.end()); while (i < nums1.size() && j < nums2.size()) { if (nums1[i] == nums2[j]) { res.push_back(nums1[i]); ++i; ++j; } else if (nums1[i] > nums2[j]) { ++j; } else { ++i; } } return res; } };
Github 同步地址：
https://github.com/grandyang/leetcode/issues/350
類似題目：
Find Common Characters
參考資料：
https://leetcode.com/problems/intersection-of-two-arrays-ii/
https://leetcode.com/problems/intersection-of-two-arrays-ii/discuss/82269/Short-Python-C%2B%2B