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 num2'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?
这道题是之前那道
Intersection of Two Arrays的拓展，不同之处在于这道题允许我们返回重复的数字，而且是尽可能多的返回，之前那道题是说有重复的数字只返回一个就行。那么这道题我们用哈希表来建立nums1中字符和其出现个数之间的映射, 然后遍历nums2数组，如果当前字符在哈希表中的个数大于0，则将此字符加入结果res中，然后哈希表的对应值自减1，参见代码如下：
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; } };
再来看一种方法，这种方法先给两个数组排序，然后用两个指针分别指向两个数组的起始位置，如果两个指针指的数字相等，则存入结果中，两个指针均自增1，如果第一个指针指的数字大，则第二个指针自增1，反之亦然，参见代码如下：
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; } };
本文转自博客园Grandyang的博客，原文链接：两个数组相交之二[LeetCode] Intersection of Two Arrays II ，如需转载请自行联系原博主。
