Javascript

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[]}
 */
var intersection = function(nums1, nums2) {
        var rec=[];
        var result=[];
        var count=0;
        for(var i=0;i<nums1.length;i++)
        {
            rec[nums1[i]]=2;
        }
        nums2.sort(function(a,b){return a-b;});
        for(var m=0;m<nums2.length-1;m++)
        {
            if(nums2[m+1]===nums2[m])
            {
            nums2.splice(m,1);
            m--;
            }
        }
        for(var j=0;j<nums2.length;j++)
        {
            if(rec[nums2[j]]==2)
            result[count++]=nums2[j];
        }
        return result;
};

C++和Java想得太复杂了，看了下别人的解答

优解

Java，用哈希表解决了重复的问题

public class Solution {
    public int[] intersection(int[] nums1, int[] nums2) {
        Set<Integer> set = new HashSet<>();
        Set<Integer> intersect = new HashSet<>();
        for (int i = 0; i < nums1.length; i++) {
            set.add(nums1[i]);
        }
        for (int i = 0; i < nums2.length; i++) {
            if (set.contains(nums2[i])) {
                intersect.add(nums2[i]);
            }
        }
        int[] result = new int[intersect.size()];
        int i = 0;
        for (Integer num : intersect) {
            result[i++] = num;
        }
        return result;
    }
}

Java，排序后用两根指针依次比对数组找出相同元素

public class Solution {
    public int[] intersection(int[] nums1, int[] nums2) {
        Set<Integer> set = new HashSet<>();
        Arrays.sort(nums1);
        Arrays.sort(nums2);
        int i = 0;
        int j = 0;
        while (i < nums1.length && j < nums2.length) {
            if (nums1[i] < nums2[j]) {
                i++;
            } else if (nums1[i] > nums2[j]) {
                j++;
            } else {
                set.add(nums1[i]);
                i++;
                j++;
            }
        }
        int[] result = new int[set.size()];
        int k = 0;
        for (Integer num : set) {
            result[k++] = num;
        }
        return result;
    }
}

C++

class Solution {
public:
    vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
    set<int> s(nums1.begin(), nums1.end());
    vector<int> out;
    for (int x : nums2)
        if (s.erase(x))
            out.push_back(x);
    return out;
    }
};