[LeetCode] Intersection of Two Arrays 两个数组的交集

声明：原题目转载自LeetCode，解答部分为原创

Problem ：

Given two arrays, write a function to compute their intersection.

Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2].

Note:

• Each element in the result must be unique.
• The result can be in any order.

Solution：

        思路：判断交集，先排序，再用两个指针分别从头到尾指向两个数组的元素。当两指针指向数值相同时，判断该数值是否存在sub容器中，方式是比较数值与sub容器的最后一位的数值是否相等，若不等，将数值push进sub容器， 并将两个指针指向下一位；当指针指向的两数值不等时，数值小的一方将指针指向下一位。重复以上操作直至其中一方数值比较完。

        代码如下：

#include<iostream>#include<vector>#include<algorithm>using namespace std;void output(vector<int> & temp){for(int i = 0 ; i < temp.size() ; i ++){cout << temp[i] << " ";}cout << endl;}class Solution {public:    vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {        sort(nums1.begin(), nums1.end());        sort(nums2.begin(), nums2.end());        vector<int> sub;                int point_1 = 0;        int point_2 = 0;        while(point_1 < nums1.size() && point_2 < nums2.size())        {        if(nums1[point_1] == nums2[point_2])        {        if(!sub.empty())        {        int pre_size = sub.size();        int sub_num = sub[pre_size - 1];        if(sub_num != nums1[point_1])        sub.push_back(nums1[point_1]);}else        sub.push_back(nums1[point_1]);        point_1 ++;        point_2 ++;        continue;        }        else if(nums1[point_1] < nums2[point_2])        {        point_1 ++;        continue;}else{point_2 ++;continue;}}return sub;    }};int main(){vector<int> text_1;vector<int> text_2;vector<int> sub;text_1.push_back(1);text_1.push_back(2);text_2.push_back(2);text_2.push_back(1);Solution text;sub = text.intersection(text_1, text_2);output(sub);return 0;}