Given two arrays, write a function to compute their intersection.
Notice
Each element in the result should appear as many times as it shows in both arrays.
The result can be in any order.
Example
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].
Challenge
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?
LeetCode上的原题，请参见我之前的博客Intersection of Two Arrays II。
解法一：
class Solution { public: /** * @param nums1 an integer array * @param nums2 an integer array * @return an integer array */ vector<int> intersection(vector<int>& nums1, vector<int>& nums2) { vector<int> res; unordered_map<int, int> m; for (auto a : nums1) ++m[a]; for (auto a : nums2) { if (m[a] > 0) { res.push_back(a); --m[a]; } } return res; } };
解法二：
class Solution { public: /** * @param nums1 an integer array * @param nums2 an integer array * @return an integer array */ vector<int> intersection(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]) ++i; else if (nums1[i] > nums2[j]) ++j; else { res.push_back(nums1[i]); ++i; ++j; } } 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[LintCode] Intersection of Two Arrays II ，如需转载请自行联系原博主。