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[LintCode] Intersection of Two Arrays 两个数组相交

更新时间:2022-09-16 19:24:38

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());
    }
};

本文转自博客园Grandyang的博客,原文链接:两个数组相交[LintCode] Intersection of Two Arrays ,如需转载请自行联系原博主。