标签：
Given two arrays, write a function to compute their intersection.
Notice
Each element in the result must be unique.
The result can be in any order.
Have you met this question in a real interview?
Example
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2].
Challenge
Can you implement it in three different algorithms?
LeetCode上的原题，请参见我之前的博客Intersection of Two Arrays。
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) { set<int> s, res; for (auto a : nums1) s.insert(a); for (auto a : nums2) { if (s.count(a)) res.insert(a); } return vector<int>(res.begin(), res.end()); } };
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 { if (res.empty() || res.back() != nums1[i]) { res.push_back(nums1[i]); } ++i; ++j; } } return res; } };
解法三：
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) { set<int> res; sort(nums2.begin(), nums2.end()); for (auto a : nums1) { if (binarySearch(nums2, a)) { res.insert(a); } } return vector<int> (res.begin(), res.end()); } bool binarySearch(vector<int> &nums, int target) { int left = 0, right = nums.size(); while (left < right) { int mid = left + (right - left) / 2; if (nums[mid] == target) return true; else if (nums[mid] < target) left = mid + 1; else right = mid; } return false; } };
解法四：
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) { set<int> s1(nums1.begin(), nums1.end()), s2(nums2.begin(), nums2.end()), res; set_intersection(s1.begin(), s1.end(), s2.begin(), s2.end(), inserter(res, res.begin())); return vector<int>(res.begin(), res.end()); } };
[LintCode] Intersection of Two Arrays 两个数组相交
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原文地址：http://www.cnblogs.com/grandyang/p/5565633.html
